PAS MATEMATIKA BLOG

Contoh 4 Soal dan Pembahasan Materi Lingkaran dan Hubungan Dua Lingkaran

$\begin{array}{ll}\\ 16.&\textrm{Salah satu garis singgung yang bersudut}\: \: 120^{\circ}\\ &\textrm{terhadap sumbu x positif terhadap lingkaran}\\ &\textrm{dengan ujung diameter titik}\: \: (7,6)\: \textrm{dan}\: \: (1,-2)\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}y=-x\sqrt{3}+4\sqrt{3}+12\\ &\textrm{b}.\quad y=-x\sqrt{3}-4\sqrt{3}+8\\ &\textrm{c}.\quad y=-x\sqrt{3}+4\sqrt{3}-4\\ &\textrm{d}.\quad y=-x\sqrt{3}-4\sqrt{3}-8\\ &\textrm{e}.\quad y=-x\sqrt{3}+4\sqrt{3}+22\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|}\hline \textrm{Pusat Lingkaran}&\textrm{Gradien Garis Singgung}\\\hline \begin{aligned}&(a,b)\\ &=\left ( \displaystyle \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2} \right )\\ &=\left ( \displaystyle \frac{7+1}{2},\frac{6+(-2)}{2} \right )\\ &=(4,2) \end{aligned}&\begin{aligned}m&=\tan 120^{\circ}\\ &=-\tan \left ( 180^{\circ}-60^{\circ} \right )\\ &=-\tan 60^{\circ}\\ &=-\sqrt{3}\\ &\\ \end{aligned} \\\hline \textrm{Jari-jari}&\textrm{Garis Singgung}\\\hline \begin{aligned}r&=\textrm{jarak titik}\\ &\: \: \: \: \: \, \textrm{singgung ke pusat}\\ &=\sqrt{(7-4)^{2}+(6-2)^{2}}\\ &=\sqrt{3^{2}+4^{2}}\\ &=\sqrt{25}\\ &=5\\ &\\ &\\ & \end{aligned}&\begin{aligned} &(y-b)=m(x-a)\pm r\sqrt{1+m^{2}}\\ &\Leftrightarrow (y-2)=-\sqrt{3}(x-4)\pm 5\sqrt{1+(-\sqrt{3})^{2}}\\ &\Leftrightarrow y-2=-\sqrt{3}x+4\sqrt{3}\pm 5\sqrt{1+4}\\ &\Leftrightarrow y=-\sqrt{3}x+4\sqrt{3}+2\pm 10\\ &\Leftrightarrow y=\begin{cases} -\sqrt{3}x+4\sqrt{3}+2+ 10 \\ -\sqrt{3}x+4\sqrt{3}+2- 10 \end{cases}\\ &\Leftrightarrow y=\begin{cases} \color{red}-\sqrt{3}x+4\sqrt{3}+12 & \\ -\sqrt{3}x+4\sqrt{3}-8 & \end{cases} \end{aligned}\\\hline \end{array}\\ &\textrm{Berikut ilustrasi gambarnya} \end{array}$.

Dengan ilustrasi tambahan

$\begin{array}{ll}\\ 17.&\textrm{Salah satu garis singgung lingkaran}\\\ & x^{2}+y^{2}=10\: \: \textrm{yang ditarik dari}\\ &\textrm{titik}\: \: (4,2)\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \color{red}x+3y=10\\ &\textrm{b}.\quad x-3y=10\\ &\textrm{c}.\quad -x-3y=10\\ &\textrm{d}.\quad 2x+y=10\\ &\textrm{e}.\quad x+2y=10\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|}\hline \begin{aligned}&\textrm{Garis Singgung}\\ &\quad\quad \textrm{di titik}\\ &(x_{1},y_{1})=(4,2) \end{aligned}&\begin{aligned}&\textrm{Tahapan menentukan}\\ &\quad\qquad \textrm{harga}\: \: m\\ & \end{aligned}\\\hline \begin{aligned}&y-y_{1}=m(x-x_{1})\\ &y-2=m(x-4)\\ &y=mx-4m+2\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&x^{2}+y^{2}=10\\ &x^{2}+\left ( mx-4m+2 \right )^{2}=10\\ &x^{2}+m^{2}x^{2}+16m^{2}+4-8m^{2}x+4mx-16m=10\\ &x^{2}+m^{2}x^{2}+16m^{2}-8m^{2}x+4mx-16m-6=0\\ &(1+m^{2})x^{2}+(4m-8m^{2})x+16m^{2}-16m-6=0\\ &\begin{cases} a & =1+m^{2} \\ b & =4m-8m^{2} \\ c & =16m^{2}-16m-6 \end{cases} \end{aligned}\\\hline \end{array}\\ &\begin{aligned}&\textrm{Syarat menyinggung}\: \: D=0\\ &b^{2}-4ac=0\\ &\left ( 4m-8m^{2} \right )^{2}-4\left ( 1+m^{2} \right )\left ( 16m^{2}-16m-6 \right )=0\\ &16m^{2}-64m^{3}+64m^{4}-64m^{2}+64m+24-64m^{4}+64m^{3}+24m^{2}=0\\ &-24m^{2}+64m+24=0\\ &-3m^{2}+8m+3=0\\ &(m-3)(3m+1)=0\\ &m=3\: \: \textrm{atau}\: \: m=-\displaystyle \frac{1}{3}\\ &m=\begin{cases} 3 & \Rightarrow y=3x-10\\ &\Rightarrow 3x-y=10\\ -\displaystyle \frac{1}{3} & \Rightarrow y=-\displaystyle \frac{1}{3}x+\frac{4}{3}+2\\ &\Rightarrow \color{red}x+3y=10 \end{cases} \end{aligned} \end{array}$.

$.\qquad\begin{aligned}&\color{purple}\textrm{Berikut ilustrasi gambarnya} \end{aligned}$

$\begin{array}{ll}\\ 18.&\textrm{Diketahui persamaan lingkaran}\: \: x^{2}+y^{2}=r^{2}\\ &\textrm{dan sebuah titik di luar lingkaran}\: \: M(a,b)\\ &\textrm{Posisi garis}\: \: ax+by=r^{2}\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \textrm{menyinggung lingkaran}\\ &\textrm{b}.\quad \color{red}\textrm{memotong lingkaran di dua titik}\\ &\textrm{c}.\quad \textrm{melalui titik pusat lingkaran}\\ &\textrm{d}.\quad \textrm{tidak memotong lingkaran}\\ &\textrm{e}.\quad \textrm{tidak ada yang benar}\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\\ &\bullet \quad L\equiv x^{2}+y^{2}=r^{2}\\ &\bullet \quad M(a,b)\: \: \textrm{di luar lingkaran}\: \: L\\ &\color{purple}\textrm{Selanjutnya perhatikan penjelasan berikut}\\ &\begin{aligned}&\textrm{Karena}\: M(a,b)\: \textrm{di luar lingkaran}\: L,\: \textrm{maka}\\ &\textrm{maka salah satu dari}\: \: a\: \: \textrm{atau}\: \: b\: \: \textrm{atau keduanya}\\ &\textrm{akan lebih besar nilanya dari pada}\: \: r.\\ &\textrm{Misalkan kita pilih}\: \: a>r\\ &\color{blue}\textrm{Ambil posisi saat memotong sumbu}-X,\: \color{black}y=0\\ &\begin{aligned}&\textrm{Untuk lingkaran}\: \: x^{2}+y^{2}=r^{2}\\ &\bullet \quad y=0\Rightarrow x^{2}+0^{2}=r^{2}\Rightarrow x=\left | r \right |\\ &\textrm{Untuk garis}\: \: ax+by=r^{2}\\ &\bullet \quad y=0\Rightarrow ax=r^{2}\Rightarrow x=\displaystyle \frac{r^{2}}{a}\\ &\textrm{Dari sini tampak posisi}\: \: x=\color{red}\left | r \right |> \displaystyle \frac{r^{2}}{a}\geq 0 \end{aligned}\\ &\textrm{Sehingga kesimpulannya adalah}:\\ &\color{red}\textrm{garis tersebut akan selalu memotong lingkaran} \end{aligned}\\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Dua lingkaran dengan persamaan}\\ &\textrm{lingkaran-lingkaran}\: x^{2}+y^{2}+6x-8y+21=0\\ &\textrm{dan}\: \: x^{2}+y^{2}+10x-8y+25=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \textrm{berpotongan di luar titik}\\ &\textrm{b}.\quad \textrm{tidak berpotongan atau bersinggungan}\\ &\textrm{c}.\quad \textrm{bersinggungan luar}\\ &\textrm{d}.\quad \color{red}\textrm{bersinggungan dalam}\\ &\textrm{e}.\quad \textrm{sepusat}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}+6x-8y+21=0&\begin{cases} P_{1} &=(-3,4) \\ r_{1} & = 2 \end{cases}\\\hline L_{2}\equiv x^{2}+y^{2}+10x-8y+25=0&\begin{cases} P_{2} &=(-5,4) \\ r_{2} & = 4 \end{cases}\\\hline \end{array} \\ &\textrm{dan}\\ &\begin{array}{|c|c|}\hline \textrm{Jarak kedua pusat}&\textrm{Jumlah/selisih jari-jari}\\\hline \begin{aligned}&\left (P_{1}P_{2} \right )\\ &=\sqrt{(-3+5)^{2}+(4-4)^{2}}\\ &=\sqrt{2^{2}+0^{2}}=\sqrt{4}=2 \end{aligned}&\begin{aligned}\begin{cases} r_{1}+r_{2} & =2+4=6 \\ \left |r_{1}-r_{2} \right | & =\left | 2-4 \right |=2 \end{cases} \end{aligned}\\\hline \end{array}\\ &\textrm{Karena nilai}\: \: \color{red}P_{1}P_{2}\color{black}=\color{red}\left |r_{1}-r_{2} \right |\color{black}=\color{red}2\\ &\textrm{hal ini menunjukkan keduanya bersinggungan}\\ &\color{blue}\textrm{di dalam}\\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut} \end{array}$ .

$\begin{array}{ll}\\ 20.&\textrm{Dua lingkaran dengan persamaan}\\ &\textrm{lingkaran-lingkaran}\: x^{2}+y^{2}+2x-6y+9=0\\ &\textrm{dan}\: \: x^{2}+y^{2}+8x-6y+9=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \textrm{berpotongan}\\ &\textrm{b}.\quad \color{red}\textrm{bersinggungan di dalam}\\ &\textrm{c}.\quad \textrm{bersinggungan luar}\\ &\textrm{d}.\quad \textrm{tidak berpotongan}\\ &\textrm{e}.\quad \textrm{sepusat}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}+2x-6y+9=0&\begin{cases} P_{1} &=(-1,3) \\ r_{1} & = 1 \end{cases}\\\hline L_{2}\equiv x^{2}+y^{2}+8x-6y+9=0&\begin{cases} P_{2} &=(-4,3) \\ r_{2} & = 4 \end{cases}\\\hline \end{array} \\ &\textrm{dan}\\ &\begin{array}{|c|c|}\hline \textrm{Jarak kedua pusat}&\textrm{Jumlah/selisih jari-jari}\\\hline \begin{aligned}&\left (P_{1}P_{2} \right )\\ &=\sqrt{(-1+4)^{2}+(3-3)^{2}}\\ &=\sqrt{3^{2}+0^{2}}=\sqrt{9}=3 \end{aligned}&\begin{aligned}\begin{cases} r_{1}+r_{2} & =1+4=5 \\ \left |r_{1}-r_{2} \right | & =\left | 1-4 \right |=3 \end{cases} \end{aligned}\\\hline \end{array}\\ &\textrm{Karena nilai}\: \: \color{red}P_{1}P_{2}\color{black}=\color{red}\left |r_{1}-r_{2} \right |\color{black}=\color{red}3\\ &\textrm{hal ini menunjukkan keduanya bersinggungan}\\ &\color{blue}\textrm{di dalam}\\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut} \end{array}$.

DAFTAR PUSTAKA

Budi, W. S. 2010. Bahan Ajar Persiapan Menuju Olimpiade Sain Nasional/Internasional Matematika 3. Jakarta: ZAMRUD KEMALA.
Kartini, Suprapto, Subandi, dan Setiadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
Kanginan M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
Noormandiri. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU
Sukino. 2017. Matematika Jilid 2 untuk Kelas SMA/MA Kelas XI Kelompok Peminatan dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh 3 Soal dan Pembahasan Materi Lingkaran

$\begin{array}{ll}\\ 11.&\textrm{Lingkaran}\: \: x^{2}+y^{2}+2ax+2by+c=0\\ &\textrm{menyinggung sumbu Y jika}\: \: c\: =....\\ &\textrm{A}.\quad ab\\ &\textrm{B}.\quad ab^{2}\\ &\textrm{C}.\quad a^{2}b\\ &\textrm{D}.\quad a^{2}\\ &\textrm{E}.\quad \color{red}b^{2}\\\\ &\textbf{Jawab}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\begin{aligned}&x^{2}+y^{2}+2ax+2by+c=0\\ &x=0\Rightarrow 0^{2}+y^{2}+2a.0+2by+c=0\\ &y^{2}+2by+c=0\begin{cases} a & =1 \\ b & =2b \\ c & =c \end{cases}\\ &\textrm{Syarat menyinggung}\: \textrm{adalah}:\\ &D=b^{2}-4ac=0\\ &\Leftrightarrow (2b)^{2}-4.1.c=0\\ &\Leftrightarrow 4c=4b^{2}\\ &\Leftrightarrow c=\color{red}b^{2} \end{aligned} \\\\ &\color{blue}\textbf{Alternatif 2}\\ &\begin{aligned}&x^{2}+y^{2}+2ax+2by+c=0\\ &\Leftrightarrow x^{2}+2ax+a^{2}+y^{2}+2by+b^{2}+c-a^{2}-b^{2}=0\\ &\Leftrightarrow (x+a)^{2}+(y+b)^{2}=a^{2}+b^{2}-c\\ &\textrm{Karena menyinggung sumbu-Y, maka}\: \: R=a \\ &\textrm{Sehingga}\: \: R^{2}=a^{2}+b^{2}-c=a^{2}\\ &\Leftrightarrow b^{2}-c=0\\ &\Leftrightarrow b^{2}=c\\ &\Leftrightarrow c=\color{red}b^{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Diketahui pusat lingkaran L terletak dikuadran}\\ &\textrm{I dan berada di sepanjang garis}\: \: y=2x.\: \: \textrm{Jika}\\ &\textrm{lingkaran L menyinggung sumbu Y di titik}\\ &(0,6),\: \textrm{maka persamaan lingkaran L adalah}\: ....\\ &\textrm{A}.\quad x^{2}+y^{2}-3x-6y=0\\ &\textrm{B}.\quad x^{2}+y^{2}+6x+12y-108=0\\ &\textrm{C}.\quad x^{2}+y^{2}+12x+6y-72=0\\ &\textrm{D}.\quad x^{2}+y^{2}-12x-6y=0\\ &\textrm{E}.\quad \color{red}x^{2}+y^{2}-6x-12y+36=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&(x-a)^{2}+(y-b)^{2}=r^{2},\\ &\textrm{menyinggung titik}\: \: (0,6)\\ &\textrm{berarti pusat lingkaran L juga terletak}\\ &\textrm{pada garis}\: \: y=6.\: \: \textrm{Hal ini menunjukkan bahwa }\\ &\textrm{pusat lingkaran}\: \: \, \: \textrm{L berpusat di}\: \: (x,2x)=(\frac{y}{2},y),\\ &\textrm{dengan}\: \: y=6.\, \: \textrm{Dari informasi di atas, }\\ &\textrm{didapatlah pusat lingkaran berada di titik}\: \: (3,6).\\ &\textrm{Sehingga persamaan lingkarannya adalah}:\\ &(x-3)^{2}+(y-6)^{2}=3^{2}\: \: \textrm{ingat}\: \: r=\textrm{absis}\: \: x=3\\ &\Leftrightarrow (x-3)^{2}+(y-6)^{2}=x^{2}-6x+9+y^{2}+12x+36=9\\ &\Leftrightarrow \, \color{red}x^{2}+y^{2}-6x+12y+36=0\\ &\color{purle}\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Persamaan garis singgung lingkaran}\\ &x^{2}+y^{2}+8x-3y-24=0,\: \: \textrm{di titik}\\ & (2,4)\: \: \textrm{adalah}\: ....\\ &\textrm{A}.\quad 12x-5y-44=0\\ &\textrm{B}.\quad \color{red}12x+5y-44=0\\ &\textrm{C}.\quad 12x-y-50=0\\ &\textrm{D}.\quad 12x+y-50=0\\ &\textrm{E}.\quad 12x+y+50=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&x^{2}+y^{2}+8x-3y-24\\ &\Leftrightarrow x^{2}+8x+16+y^{2}-3y+\displaystyle \frac{9}{4}-24=16+\frac{9}{4}\\ &\Leftrightarrow \: (x+4)^{2}+(y-\frac{3}{2})^{2}=16+\frac{9}{4}+24=42\frac{1}{4}\\ &\textrm{Persamaan garis singgung lingkar}\textrm{an lingkaran }\\ &\textrm{di titik}\: \: (x_{1},y_{1})\: \: \textrm{adalah}:\\ &(x_{1}+4)(x+4)+(y_{1}-\frac{3}{2})(y-\frac{3}{2})=42\frac{1}{4},\\ &\textrm{untuk}\: \: (x_{1},y_{1})=(2,4),\: \textrm{maka}\\ &(2+4)(x+4)+(4-\frac{3}{2})(y-\frac{3}{2})=\frac{169}{4}\\ &\Leftrightarrow 6(x+4)+\frac{5}{2}(y-\frac{3}{2})=\frac{169}{4}\\ &\Leftrightarrow 24(x+4)+5(2y-3)=169\\ &\Leftrightarrow 24x+96+10y-15=169\\ &\Leftrightarrow 24x+10y=169-96+15=88\\ &\Leftrightarrow \color{red}12x+5y-44=0\\ &\color{purple}\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Sebuah garis singgung}\: \: g\: \: \textrm{menyinggung }\\ &\textrm{lingkaran yang berpusat di}\: \: (-2,5)\: \: \textrm{dan}\\ &\textrm{berjari-jari}\: \: 2\sqrt{10}\: \: \textrm{di titk}\: \: (4,3),\: \textrm{maka }\\ &\textrm{persamaan garis singgung}\: \: g\: \: \textrm{adalah}\: .... \\ &\textrm{A}.\quad y=3x+9\\ &\textrm{B}.\quad \color{red}y=3x-9\\ &\textrm{C}.\quad y=-3x+9\\ &\textrm{D}.\quad y=-3x-9\\ &\textrm{E}.\quad y=3x+21\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&(x-a)^{2}+(y-b)^{2}=r^{2}\\ &\begin{cases} \textrm{Pusat} & =(-2,5) \\ \textrm{r} & =2\sqrt{10} \end{cases} \\ &\textrm{maka persamaan lingkarannya}:\\ &(x+2)^{2}+(y-5)^{2}=(2\sqrt{10})^{2}\\ &\Leftrightarrow (x_{1}+2)(x+2)+(y_{1}-5)(y-5)=40,\\ &\textrm{menyingung garis}\: \: g\: \: \textrm{di}\: (4,3)\\ &(4+2)(x+2)+(3-5)(y-5)=40\\ &\Leftrightarrow 6x+12-2y+10=40\\ &\Leftrightarrow 6x-2y=40-12-10\\ &\Leftrightarrow 3x-y=9\\ &\Leftrightarrow -y=-3x+9\\ &\Leftrightarrow \color{red}y=3x-9\\ &\color{purple}\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Suatu lingkaran dengan titik pusatnya terletak }\\ &\textrm{pada kurva}\: \: y=\sqrt{x}\: \: \textrm{dan melalui titik asal}\: \: O(0,0).\\ & \textrm{Jika diketahui absis titik pusat lingkaran tersebut }\\ &\textrm{adalah}\: \: a,\: \: \textrm{maka persamaan garis singgung }\\ &\textrm{lingkaran yang melalui titik}\: \: O\: \: \textrm{tersebut adalah}\: ....\\ &\textrm{A}.\quad y=-x\\ &\textrm{B}.\quad \color{red}y=-x\sqrt{a}\\ &\textrm{C}.\quad y=-ax\\ &\textrm{D}.\quad y=-2x\sqrt{2}\\ &\textrm{E}.\quad y=-2ax\\\\ &\textbf{Jawab}:\\ &\begin{array}{|l|c|l|}\hline \begin{aligned}&\textrm{Pusat}\\ &\textrm{lingkaran}\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{Gradien garis singgung}\\ &\textrm{yang tegak lurus dengan }\\ &\textrm{garis yang melalui titik}\\ &\textrm{pusat lingkaran yang }\\ &\textrm{bergradien}\: \: m_{L} \end{aligned}&\begin{aligned}&\textrm{Persamaan garis }\\ &\textrm{singgung yang }\\ &\textrm{melalui titik asal}\\ &O(0,0)\\ & \end{aligned}\\\hline \begin{aligned}&(a,b)\\ &=\left ( a,\sqrt{a} \right )\\ &\\ &\\ & \end{aligned}&\color{blue}\begin{aligned}&m.m_{1}=-1\\ &m.\frac{y}{x}=-1\\ &m=-\frac{x}{y}=-\displaystyle \frac{a}{\sqrt{a}}\\ &\: \: \: \, =-\sqrt{a} \end{aligned}&\begin{aligned}y&=mx,\\ & \textrm{karena melalui}\\ &\textrm{titik asal}\\ y&=-\sqrt{a}x,\\ y&=\color{red}-x\sqrt{a} \end{aligned}\\\hline \end{array} \end{array}$.

Contoh 2 Soal dan Pembahasan Materi Lingkaran

$\begin{array}{ll}\\ 6.&\textrm{Diketahui lingkaran}\: \: x^{2}+y^{2}+4x+ky-12=0\\ &\textrm{melalui titik}\: \: (-2,8)\: \: \textrm{maka jari-jari lingkaran}\\ &\textrm{tersebut adalah}....\\ &\textrm{A}.\quad 1\\ &\textrm{B}.\quad \color{red}5\\ &\textrm{C}.\quad 6\\ &\textrm{D}.\quad 12\\ &\textrm{E}.\quad 25\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui ingkaran berpusat di}\: \left ( -2,-\displaystyle \frac{1}{2}k \right ),\\ &\textrm{yaitu}:\\ &x^{2}+y^{2}+4x+ky-12=0\\ & \textrm{melalui}\: \: (-2,8)\: \: \textrm{berarti }\\ &(-2)^{2}+8^{2}+4(-2)+k.8-12=0\\ &4+64-8-12+8k=0\\ &48+8k=0\\ &k=\color{blue}-6\\ &\textrm{Sehingga}\: \: r=\sqrt{\displaystyle \frac{4^{2}}{4}+\frac{(-6)^{2}}{4}-(-12)}\\ &\qquad\qquad \: \: \: =\sqrt{\displaystyle 4+9+12}=\sqrt{25}=\color{red}5\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Persmaan lingkaran}\: \: x^{2}+y^{2}+px+8y+9=0\\ &\textrm{menyinggung sumbu X. Pusat lingkaran tersebut }\\ &\textrm{adalah}\: ....\\ &\textrm{A}.\quad (6,-4)\\ &\textrm{B}.\quad (6,6)\\ &\textrm{C}.\quad \color{red}(3,-4)\\ &\textrm{D}.\quad (-6,-4)\\ &\textrm{E}.\quad (3,4)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textbf{Lingkaran}\: \: x^{2}+y^{2}+px+8y+9=0\\ &\textrm{maka,}\\ &x^{2}+px+y^{2}+8y+9=0\\ &\left ( x+\displaystyle \frac{1}{2}p \right )^{2}-\displaystyle \frac{1}{4}p^{2}+(y+4)^{2}-16+9=0\\ &\Leftrightarrow \left ( x+\displaystyle \frac{1}{2}p \right )^{2}+(y+4)^{2}=7+\displaystyle \frac{1}{4}p^{2}\\ &\textrm{karena menyinggung sumbu-X,}\: \: \: \: R=b=4,\\ & \textrm{sehingga}\\ &7+\displaystyle \frac{1}{4}p^{2}=4^{2}\\ &\Leftrightarrow \displaystyle \frac{1}{4}p^{2}=16-7=9\Leftrightarrow p^{2}=36\Leftrightarrow p=\color{blue}\pm 6\\ &p=-6\: \Rightarrow \: x^{2}+y^{2}-6x+8y+9=0\\ &\quad\Rightarrow \textrm{pusatnya adalah}\: \: \left ( -\displaystyle \frac{A}{2},-\frac{B}{2} \right )=\color{red}(3,-4)\\ &p=6\: \: \: \, \: \Rightarrow \: x^{2}+y^{2}+6x+8y+9=0\\ &\quad\Rightarrow \textrm{pusatnya adalah}\: \: \left ( -\displaystyle \frac{A}{2},-\frac{B}{2} \right )=\color{red}(-3,-4)\\ &\color{purple}\textrm{dan berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Titik-titik berikut yang posisinya berada di luar }\\ &\textrm{lingkaran}\: \: x^{2}+y^{2}-2x+8y-32=0\: \: \textrm{adalah}.... \\ &\textrm{A}.\quad (0,0)\\ &\textrm{B}.\quad (-6,-4)\\ &\textrm{C}.\quad \color{red}(-3,2)\\ &\textrm{D}.\quad (3,1)\\ &\textrm{E}.\quad (4,1)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\begin{array}{|c|c|l|c|}\hline \color{blue}\textrm{Opsi}&\color{blue}\textrm{Titik}&\qquad\qquad\quad\color{blue}\textrm{Lingkaran}&\color{blue}\textrm{Keterangan}\\\hline \textrm{A}&(0,0)&0^{2}+0^{2}-2.0+8.0-32=-32&\textrm{dalam}\\\hline \textrm{B}&(-6,-4)&(-6)^{2}+(-4)^{2}-2(-6)+8(-4)-32=0&\textrm{pada}\\\hline \color{red}\textrm{C}&(-3,2)&(-3)^{2}+(2)^{2}-2(-3)+8(2)-32=3&\textbf{di luar}\\\hline \textrm{D}&(3,1)&3^{2}+1^{2}-2.3+8.1-32=-20&\textrm{dalam}\\\hline \textrm{E}&(4,1)&4^{2}+1^{2}-2.4+8.1-32=-15&\textrm{dalam}\\\hline \end{array} \\ &\color{purple}\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Diketahui garis}\: \: x-2y=5\: \: \textrm{memotong lingkaran}\\ &x^{2}+y^{2}-4y+8y+10=0\: \: \textrm{di titik A dan B}.\\ &\textrm{Panjang ruas garis AB adalah}....\\ &\textrm{A}.\quad 4\sqrt{2}\\ &\textrm{B}.\quad \color{red}2\sqrt{5}\\&\textrm{C}.\quad \sqrt{10}\\ &\textrm{D}.\quad 5\\ &\textrm{E}.\quad 4\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\begin{aligned}&\textrm{Perhatikanlah bahwa garis}\: \: \color{blue}x-2y=5\\&\textrm{memotong lingkaran}\\ &x^{2}+y^{2}-4x+8y+10=0,\\ &\textrm{maka garis}\: \: \color{blue}x=2y+5\: \: \color{black}\textrm{disubstitusikan ke}\\ &\textrm{lingkaran tersebut, yaitu}:\\ &(\color{blue}2y+5\color{black})^{2}+y^{2}-4(\color{blue}2y+5\color{black})+8y+10=0\\ &4y^{2}+20y+25+y^{2}-8y-20+8y+10=0\\ &5y^{2}+20y+15=0\\ &y^{2}+4y+3=0\\ &(y+1)(y+3)=0\\ &y=-1\: \: \vee \: \: y=-3\\ &\textrm{untuk nilai}\\ & y=-3\Rightarrow x=2(-3)+5=-1,\quad A(-1,-3)\\ &y=-1\Rightarrow x=2(-1)+5=3,\qquad B(3,-1)\\ &\textrm{maka},\qquad \textrm{AB}=\sqrt{(3-(-1))^{2}+(-1-(-3))^{2}}\\ &=\sqrt{4^{2}+2^{2}}\\ &=\sqrt{16+4}\\ &=\sqrt{20}\\ &=\color{red}2\sqrt{5} \end{aligned}\\ &\color{purple}\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$ .

$\begin{array}{ll}\\ 10.&\textrm{Kekhususan persamaan lingkaran}\\ &x^{2}+y^{2}-6x-6y+6=0\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \textrm{menyinggung sumbu X}\\ &\textrm{B}.\quad \textrm{menyinggung sumbu Y}\\ &\textrm{C}.\quad \textrm{berpusat di}\: \: O(0,0)\\ &\textrm{D}.\quad \color{red}\textrm{titik pusatnya terletak pada}\: \: x-y=0\\ &\textrm{E}.\quad \textrm{berjari-jari 3}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui persamaan lingkaran}\\ &x^{2}+y^{2}-6x-6y+6=0\\ &x^{2}-6x+9+y^{2}-6y+9+6=9+9\\ &(x-3)^{2}+(y-3)^{2}=18-6\\ &(x-3)^{2}+(y-3)^{2}=12\\ &(x-3)^{2}+(y-3)^{2}=\left ( 2\sqrt{3} \right )^{2}\\ &\textrm{lingkaran ini}\begin{cases} \textrm{Pusat} &=\color{blue}(3,3) \\ \textrm{Jari-jari} &=\color{blue}2\sqrt{3} \end{cases}\\ &\begin{array}{|c|l|c|}\hline \textrm{Opsi}&\qquad\qquad\qquad\textrm{Pernyataan}&\textrm{Keterangan}\\\hline \textrm{A}&\textrm{menyinggung sumbu X}&\textrm{tidak tepat}\\\hline \textrm{B}&\textrm{menyinggung sumbu Y}&\textrm{tidak tepat}\\\hline \textrm{C}&\textrm{berpusat di}\: \: O(0,0)&\textrm{tidak tepat}\\\hline \color{red}\textrm{D}&\color{red}\textrm{titik pusatnya terletak pada garis}\: \: x-y=0&\textbf{tepat}\\\hline \textrm{E}&\textrm{berjari-jari 3}&\textrm{tidak tepat}\\\hline \end{array} \\ &\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

Contoh 1 Soal dan Pembahasan Materi Lingkaran

$\begin{array}{ll}\\ 1.&\textrm{Jari-jari lingkaran dengan persamaan}\: \: x^{2}+y^{2}=48\\ &\textrm{adalah}....\\ &\textrm{A}.\quad \displaystyle 3\sqrt{5}\\ &\textrm{B}.\quad \color{red}4\sqrt{3}\\ &\textrm{C}.\quad 5\sqrt{2}\\ &\textrm{D}.\quad \displaystyle 6\sqrt{3}\\ &\textrm{E}.\quad 7\\\\ &\textbf{Jawab}:\qquad \\ &\begin{aligned}r^{2}&=48\\ r&=\sqrt{48}\\ &=\sqrt{16.3}\\ &=4\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Titik pusat lingkaran}\: \: (x-7)^{2}+(y+9)^{2}=48\\ &\textrm{adalah}....\\ &\textrm{A}.\quad \displaystyle (-7,-9)\\ &\textrm{B}.\quad (-7,9)\\ &\textrm{C}.\quad \color{red}(7,-9)\\ &\textrm{D}.\quad \displaystyle (7,6)\\ &\textrm{E}.\quad (15,48)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Jelas bahwa}\: \: \: (a,b)&=(-6,9) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Persamaan lingkaran yang berpusat di}\: \: P(-2,5)\\ &\textrm{dan melalui titik}\: \: T(3,4)\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \color{red}(x+2)^{2}+(y-5)^{2}=26\\ &\textrm{B}.\quad (x-3)^{2}+(y+5)^{2}=36\\ &\textrm{C}.\quad (x+2)^{2}+(y-5)^{2}=82\\ &\textrm{D}.\quad (x-3)^{2}+(y+5)^{2}=82\\ &\textrm{E}.\quad (x+2)^{2}+(y+5)^{2}=82\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Persamaan Lingkaran Berpusat di}\: \: (a,b)\\ & \textrm{adalah}:\: (x-a)^{2}+(y-b)^{2}=r^{2}\\ &\begin{array}{|l|l|l|}\hline \textrm{Pusat di}\: \: P(-2,5)&\textrm{Melalui Titik}\: \: T(3,4)\\\hline \begin{aligned}(x-a)^{2}+(y-b)^{2}&=r^{2}\\ (x+2)^{2}+(y-5)^{2}&=r^{2}\\ &\\ & \end{aligned}&\begin{aligned}(x-a)^{2}+(y-b)^{2}&=r^{2}\\ (3+2)^{2}+(4-5)^{2}&=r^{2}\\ 5^{2}+(-1)^{2}&=r^{2}\\ 26&=r^{2} \end{aligned}\\\hline \begin{aligned}&\textrm{Sehinga persamaan}\\ &\textrm{lingkarannya} \end{aligned}&\begin{aligned}&\textrm{adalah}:\\ &(x+2)^{2}+(y-5)^{2}=r^{2}=26\\ &(x+2)^{2}+(y-5)^{2}=26\\ & \end{aligned}\\\hline \end{array} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Koordinat titik pusat dan jari-jari lingkaran}\: \: x^{2}+y^{2}-4x+6y+4=0\: \: \textrm{adalah}....\\ &\textrm{A}.\quad (-3,2)\: \: \textrm{dan}\: \: 3\\ &\textrm{B}.\quad (3,-2)\: \: \textrm{dan}\: \: 3\\ &\textrm{C}.\quad (-2,-3)\: \:\textrm{ dan}\: \: 3\\ &\textrm{D}.\quad \color{red}(2,-3)\: \: \textrm{dan}\: \: 3\\ &\textrm{E}.\quad (2,3)\: \: \textrm{dan}\: \: 3\\\\ &\textbf{Jawab}: \\ &\textbf{Alterntif 1}\\ &\begin{array}{|l|l|}\hline &{\textrm{Persamaan Lingkaran Berpusat di}\: \: (a,b)\: \: \textrm{dan berjari-jari}\: \: r\: \: \textrm{adalah}}\\ &{\begin{aligned}(x-a)^{2}+(y-b)^{2}&=r^{2}\\ x^{2}+y^{2}-4x+6y+4&=0\\ x^{2}-4x+y^{2}+6y+4&=0\\ x^{2}-4x+4-4+y^{2}+6y+9-9+4&=0\\ (x-2)^{2}-4+(y+3)^{2}-9+4&=0\\ (x-2)^{2}+(y+3)^{2}&=4+9-4\\ (x-2)^{2}+(y+3)^{2}&=9\\ (x-2)^{2}+(y-(-3))^{2}&=3^{2}\begin{cases} \textrm{Pusat} & =(2,-3) \\ \textrm{dan}\\ \: r & = 3 \end{cases} \end{aligned}}\\\hline \end{array}\\ &\textbf{Alterntif 2}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{persamaan lingkaran}:\: \: x^{2}+y^{2}-4x+6y+4=0\begin{cases} A & =-4 \\ B & =6 \\ C & =4 \end{cases}\\ &x^{2}+y^{2}+Ax+By+C=0\\ &\begin{cases} \textrm{Pusat} & =\left ( -\displaystyle \frac{1}{2}A,\: -\frac{1}{2}B \right )=\left ( -\frac{1}{2}\cdots ,\: -\frac{1}{2}\cdots \right )=(\cdots ,\cdots ) \\ \textrm{Jari-jari} & =\sqrt{\displaystyle \frac{1}{4}A^{2}+\frac{1}{4}B^{2}-C}=\sqrt{\displaystyle \frac{1}{4}\cdots ^{2}+\frac{1}{4}\cdots ^{2}-\cdots }=\sqrt{\cdots } \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Suatu lingkaran}\: \: x^{2}+y^{2}-4x+2y+p=0\\ &\textrm{berjari-jari 3, maka nilai}\: \: p\: \: \textrm{adalah}....\\ &\textrm{A}.\quad -1\\ &\textrm{B}.\quad -2\\ &\textrm{C}.\quad -3\\ &\textrm{D}.\quad \color{red}-4\\ &\textrm{E}.\quad -5\\\\ &\textbf{Jawab}:\\ &\begin{aligned}r=\sqrt{\displaystyle \frac{A^{2}}{4}+\frac{B^{2}}{4}-C}&=3\\ \displaystyle \sqrt{\frac{(-4)^{2}}{4}+\frac{2^{2}}{4}-p}&=3\\ \displaystyle \frac{16}{4}+\frac{4}{4}-p&=9\\ 4+1-p&=9\\ -p&=9-5\\ p&=-4 \end{aligned} \end{array}$.

Hubungan Dua Lingkaran

Hubungan Dua Buah Lingkaran

Coba perhatikan ilustrasi beberapa lingkaran berikut

Sebagai penjelasan dari kondisi di atas adalah:

$\begin{array}{|c|c|l|}\hline \textbf{Kedudukan}&\textbf{Ilustrasi}&\qquad\qquad\: \textbf{Keterangan}\\\hline \left | L_{1}L_{2} \right |>r_{1}+r_{2}&\textbf{Gambar 1}&\begin{aligned}&\textrm{kedua lingkaran tidak berpotongan}\\ &\textrm{dan tidak pula bersinggungan}\\ &\textrm{dan saling lepas} \end{aligned}\\\hline \left | L_{1}L_{2} \right |=0&\textbf{Gambar 5}&\textrm{Dikarenakan sepusat}\\\hline \left | L_{1}L_{2} \right |\leq r_{1}+r_{2}&\textbf{Gambar 6}&\textrm{Terletak di dalam lingkaran}\: \: L_{1} \\\hline \left | L_{1}L_{2} \right |=r_{1}+r_{2}&\textbf{Gambar 2}&\begin{aligned}&\textrm{kedua lingkaran tidak berpotongan}\\ &\textrm{tetapi bersinggungan di luar} \end{aligned}\\\hline \left | L_{1}L_{2} \right |=r_{1}-r_{2}&\textbf{Gambar 3}&\begin{aligned}&\textrm{kedua lingkaran tidak berpotongan}\\ &\textrm{tetapi bersinggungan di dalam} \end{aligned}\\\hline \begin{cases} \left | L_{1}L_{2} \right | > r_{1}-r_{2} \\ \left | L_{1}L_{2} \right | < r_{1}+r_{2} \end{cases}&\textbf{Gambar 4}&\begin{aligned}&\textrm{kedua lingkaran berpotongan} \end{aligned}\\\hline \end{array}$.

$\begin{aligned}&\textbf{Kuasa}\\ &\begin{array}{|l|l|}\hline \textrm{Lingkaran}&\textrm{Posisi sebuah titik terhadap lingkaran}\\\hline \begin{aligned}&\textrm{Titik dua}\\ & \textrm{lingkaran} \end{aligned}&\begin{aligned}&\textrm{Tempat kedudukan titik-titik yang memiliki}\\ &\textrm{kuasa yang sama terhadap dua lingkaran} \end{aligned}\\\hline \begin{aligned}&\textrm{Garis tiga}\\ & \textrm{lingkaran} \end{aligned}&\begin{aligned}&\textrm{Tempat kedudukan titik yang memiliki}\\ &\textrm{kuasa yang sama terhadap tiga buah lingkaran} \end{aligned}\\\hline \end{array} \end{aligned}$.

$\begin{aligned}&\textbf{Berkas Lingkaran}\\ &\begin{array}{|l|l|l|}\hline \textrm{Istilah}&\textrm{Posisi}&\qquad\qquad\textrm{Keterangan}\\\hline \begin{aligned}&\textrm{Berkas}\\ &\textrm{Lingkaran} \end{aligned}&\begin{aligned}&\textrm{Pada garis}\\ &\textrm{busur} \end{aligned}&\begin{aligned}&\textrm{Sejumlah lingkaran yang dapat }\\ &\textrm{dibuat melalui titik-titik potong }\\ &\textrm{kedua lingakaran itu}\end{aligned} \\\hline \end{array} \end{aligned}$.

$\begin{aligned}&\textbf{Tali Busur Sekutu}\\ &\begin{array}{|l|l|l|}\hline \textrm{Istilah}&\textrm{Posisi}&\qquad\qquad\textrm{Keterangan}\\\hline \begin{aligned}&\textrm{Tali Busur}\\ &\textrm{Sekutu}\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{Kedua}\\ &\textrm{lingkaran}\\ &\textrm{yang}\\ & \textrm{berpotongan} \end{aligned}&\begin{aligned}&\textrm{Ruas garis yang menghubungkan }\\ &\textrm{titik-titik potong irisan irisan }\\ &\textrm{kedua lingkaran tersebut}\\ & \end{aligned} \\\hline \end{array}\\ &\bullet \: \: \textrm{Persamaan Tali Busur Sekutunya adalah}:\: \color{blue}L_{1}-L_{2}=0\\ &\bullet \: \: \textrm{Persamaan yang melalui titik potong dan lingkaran (berkas)}\\ &\: \: \quad \textrm{itu adalah}:\: L_{3}=L_{1}+\color{red}p\color{black}(L_{1}-L_{2}),\: \: \textrm{atau}\: \: L_{3}=L_{1}+\color{red}p\color{black}L_{2}\\ &\: \: \quad \textrm{dengan}\: \: \color{red}p\: \: \color{black}\textrm{adalah suatu parameter (suatu patokan nilai)}\\ &\bullet \: \: \textrm{Luas daerah irisan}:\: (\theta _{1}r_{1}^{2}+\theta _{2}r_{2}^{2})-\displaystyle \frac{1}{2}(r_{1}^{2}\sin \theta _{1}+r_{2}^{2}\sin \theta _{2}) \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah kedudukan untuk dua buah lingkaran}\\ & L_{1}\equiv x^{2}+y^{2}-2x-4y+1=0\\ &\textrm{dan}\: \: L_{2}\equiv x^{2}+y^{2}-4x-2y-1=0.\\ & \textrm{Jika kedua lingkaran tersebut bersinggungan}\\ &\textrm{atau berpotongan, tentukanlah titik singgung atau potongnya} \\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|}\hline L_{1}&L_{2}\\\hline x^{2}+y^{2}-2x-4y+1=0&x^{2}+y^{2}-4x-2y-1=0\\\hline \begin{cases} P_{1} & :\left ( - \displaystyle \frac{1}{2}(-2),-\frac{1}{2}(-4) \right )=(1,2) \\ r &=\sqrt{\displaystyle \frac{1}{4}\left ( (-2)^{2}+(-4^{2}) \right )-1}\\ &=2 \end{cases}&\begin{cases} P_{2} & :\left ( - \displaystyle \frac{1}{2}(-4),-\frac{1}{2}(-2) \right )=(2,1) \\ r &= \sqrt{\displaystyle \frac{1}{4}\left ( (-4)^{2}+(-2^{2}) \right )-(-1)}\\ &=\sqrt{6} \end{cases}\\\hline \end{array}\\ &\begin{aligned}\textrm{Jarak ke}&\textrm{dua pusat lingkarannya adalah}\: \: P_{1}P_{2}\: \: \textrm{yaitu}:\\ P_{1}P_{2}&=\sqrt{(2-1)^{2}+(1-2)^{2}}\\ &=\sqrt{2}\\ \textrm{Karena n}&\textrm{ilai}\: \: P_{1}P_{2}=\sqrt{2}\: \: \textrm{dan nilai}\: \: P_{1}+P_{2}=2+\sqrt{6},\\ \textrm{sehingga}\: &P_{1}P_{2}<P_{1}+P_{2}\: \: \textrm{maka kedua lingkaran }\\ \textrm{itu berpo}&\textrm{tongan} \end{aligned} \end{array}$.

$.\qquad \begin{aligned}x^{2}+y^{2}-2x-4y+1&=0\: ..................(1)\\ x^{2}+y^{2}-4x-2y-1&=0\: ..................(2)\\ ----------&---\: ^{-}\\ 2x-2y+2&=0\\ y&=x+1\: ........................(3)\\ \textrm{persamaan}\: \: (3)\rightarrow (1)&\\ x^{2}+(x+1)^{2}-2x-&4(x+1)+1=0\\ x^{2}+x^{2}+2x+1-2x&-4x-4+1=0\\ 2x^{2}-4x-2&=0\Leftrightarrow x^{2}-2x-1=0\\ x_{1,2}&=\displaystyle \frac{-(-2)\pm \sqrt{(-2)^{2}-4.1(-1)}}{2}\\ &=\displaystyle \frac{2\pm \sqrt{8}}{2}=\displaystyle \frac{2\pm 2\sqrt{2}}{2}\begin{cases} x_{1} & =1+\sqrt{2}\: .........(4)\: \: \textbf{atau} \\ x_{2} & =1-\sqrt{2}\: .........(5) \end{cases}\\ \textrm{persamaan}\: \: (4)\rightarrow (3)&,\: y_{1}=1+\sqrt{2}+1=2+\sqrt{2}\\ \textrm{persamaan}\: \: (5)\rightarrow (3)&,\: y_{1}=1-\sqrt{2}+1=2-\sqrt{2}\\ \textrm{Sehingga titik poton}&\textrm{gnya ada 2 yaitu}:\\ &\color{red}\begin{cases} \left ( 1+\sqrt{2},2+\sqrt{2} \right )\: \: \textrm{dan} \\ \left ( 1-\sqrt{2},2-\sqrt{2} \right ) \end{cases}& \\\textrm{Berikut ilustrasinya} \end{aligned}$.

$\begin{array}{ll}\\ 2&\textrm{Dari contoh soal no.1, tentukanlah persamaan lingkaran }\\ &\textrm{yang melalui titik potong kedua lingkaran itu serta }\\ &\textrm{melalui titik pusat koordinat}\: \: O(0,0)\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Pada jawaban soal no.1 didapatkan persamaan }\\ &\textbf{tali busur}:L_{1}-L_{2}\equiv x-y+1=0\\ &\textrm{Sehingga persamaan }\\ &\textbf{berkas lingkaran}\textrm{nya adalah}:L_{3}=L_{1}+p\left (L_{1}-L_{2} \right )=0\\ &\, \, \: \qquad \Leftrightarrow L_{3}= \left (x^{2}+y^{2}-2x-4y+1 \right )+p(x-y+1)=0\\ &\textrm{Karena melalui titik asal}\: \: O(0,0),\: \textrm{maka}\\ &\, \, \: \qquad \Leftrightarrow (0+0-0-0+1)+p(0-0+1)=0\Leftrightarrow p=-1\\ &\textrm{Selanjutnya persamaan berkas lingkarannya akan menjadi}\\ &\, \, \: \qquad L_{3}\equiv x^{2}+y^{2}-2x-4y+1 -(x-y+1)=0\\ &\textrm{Jadi},\: L_{3}\equiv x^{2}+y^{2}-3x-3y=0 \\\\ &\textrm{Dan gambar berikut sebagai ilustrasinya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3&\textrm{Diketahuin dua buah lingkaran}\\ & L_{1}\equiv x^{2}+y^{2}-15y+32=0\quad \: \textrm{dan}\\ & L_{2}\equiv x^{2}+y^{2}-18x+2y+32=0\\ &\textrm{Tunjukkan bahwa kedua lingkaran}\\ &\textrm{bersinggungan di luar dan tentukan}\\ &\textrm{titik singgungnya}\\\\ &\textbf{Jawab}:\\ &\color{blue}\textrm{Akan ditunjukkan kedua lingkaran saling}\\ &\textrm{bersinggungan di luar, yaitu:}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}-15y+32=0&\begin{cases} P_{1} &=(0,8) \\ r_{1} & = 4\sqrt{2} \end{cases}\\\hline L_{2}\equiv x^{2}+y^{2}-18x+2y+32=0&\begin{cases} P_{2} &=(9,-1) \\ r_{2} & = 5\sqrt{2} \end{cases}\\\hline \end{array}\\ &\textrm{Selanjutnya}\\ &\begin{array}{|l|l|}\hline \textrm{Hitungan jarak kedua pusat}&\textrm{Sebagai bandingan}\\\hline \begin{aligned}&\textrm{Pusat 1 lingkaran}\: P_{1}=(0,8)\\ &\textrm{Pusat 2 lingkaran}\: P_{2}=(9,-1)\\ &\textrm{maka jarak}\: \: \: P_{1}P_{2}\: \: \textrm{adalah}\\ &=\sqrt{(9-0)^{2}+(-1-8)^{2}}\\ &=\sqrt{9^{2}+9^{2}}=\sqrt{2\times 9^{2}}=\color{red}9\sqrt{2} \end{aligned}&\begin{aligned}P_{1}P_{2}&=r_{1}+r_{2}\\&=4\sqrt{2}+5\sqrt{2}\\&=\color{red}9\sqrt{2}\\ &\\ & \end{aligned}\\\hline \end{array}\\ &\color{blue}\textrm{Adapun koordinat titik singgungnya}:\\ &\begin{aligned}\begin{pmatrix} x\\ y \end{pmatrix}&=\displaystyle \frac{5\begin{pmatrix} 0\\ 8 \end{pmatrix}+4\begin{pmatrix} 9\\ -1 \end{pmatrix}}{5+4}=\displaystyle \frac{\begin{pmatrix} 5\times 0+4\times 9\\ 5\times 8+4\times (-1) \end{pmatrix}}{9}\\ &=\displaystyle \frac{\begin{pmatrix} 36\\ 36 \end{pmatrix}}{9}=\begin{pmatrix} 4\\ 4 \end{pmatrix} \end{aligned}\\ &\textrm{Sehingga koordinat titik potongnya adalah}:\: (4,4)\\ &\textbf{Sebagai gambaran perhatikan ilustrasi berikut} \end{array}$.

DAFTAR PUSTAKA

Kanginan M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
Noormandiri. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU
Sukino. 2017. Matematika Jilid 2 untuk Kelas SMA/MA Kelas XI Kelompok Peminatan dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh 7 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 31.&\textrm{Diketahui}\: \: \left | \vec{a} \right |=\sqrt{3}\: ,\left | \vec{b} \right |=1,\: \: \textrm{dan}\: \: \left | \vec{a}-\vec{b} \right |=1\\ &\textrm{maka panjang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{3}&&&\textrm{d}.&2\sqrt{2}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\color{red}\sqrt{7}&\textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\: \, \textrm{sebagaimana pada soal}\\ \left | \vec{a}-\vec{b} \right |^{2}&=\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}-2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ 1^{2}&=\left ( \sqrt{3} \right )^{2}+1^{2}-2.\sqrt{3}.1.\cos \theta \\ 2\sqrt{3}\cos \theta &=3\\ \textrm{maka pan}&\textrm{jang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\\ \left | \vec{a}+\vec{b} \right |&=\sqrt{\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta}\\ &=\sqrt{\left ( \sqrt{3} \right )^{2}+1^{2}+3}\\ &=\sqrt{3+1+3}\\ &=\color{red}\sqrt{7} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Jika}\: \: \left | \overline{u} \right |=6\: ,\: \left | \overline{v} \right |=4\sqrt{3},\: \: \textrm{dan}\: \: \left | \overline{u}-\overline{v} \right |=8\\ &\textrm{tentukanlah nilai dari}\\ &\textrm{a}.\quad \overline{u}\bullet \overline{v}\\ &\textrm{b}.\quad \left | \overline{u}+\overline{v} \right |\\ &\textrm{c}.\quad \textbf{cosinus}\: \: \textrm{sudut antara}\: \: \overline{u}\: \: \textrm{dan}\: \: \overline{v}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad &\overline{u}\bullet \overline{v}=\: \cdots \\ &\: 2.\overline{u}\bullet \overline{v}=\left | \overline{u} \right |^{2}+\left | \overline{v} \right |^{2}-\left | \overline{u}-\overline{v} \right |^{2}\\ &\: 2.\overline{u}\bullet \overline{v}=6^{2}+(4\sqrt{3})^{2}-8^{2}\\ &\: 2. \overline{u}\bullet \overline{v}=36+48-64=84-64=20\\ &\quad \overline{u}\bullet \overline{v}=\displaystyle \frac{20}{2}=\color{red}10\\ \textrm{b}.\quad &\left | \overline{u}+\overline{v} \right |^{2}=\left | \overline{u} \right |^{2}+\left | \overline{v} \right |^{2}+2.\overline{u}\bullet \overline{v}\\ &\left | \overline{u}+\overline{v} \right |^{2}=6^{2}+(4\sqrt{3})^{2}+20\\ &\: \: \quad\qquad =84+20=104\\ &\left | \overline{u}+\overline{v} \right |=\color{red}\sqrt{104}\\ \textrm{c}.\quad &\cos \angle (\overline{u},\, \overline{v})=\displaystyle \frac{\overline{u}\bullet \overline{v}}{\left |\overline{u} \right |.\left | \overline{v} \right |}=\frac{10}{6.(4\sqrt{3})}\times \frac{\sqrt{3}}{\sqrt{3}}\\ &\quad\qquad\qquad =\displaystyle \frac{10\sqrt{3}}{72}=\color{red}\frac{5}{36}\sqrt{3}\\ &\color{blue}\textbf{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 33.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} 2\\ -5 \end{pmatrix},\: \vec{q}=\begin{pmatrix} 4\\ 3 \end{pmatrix},\: \textrm{maka }\\ &\textrm{proyeksi skalar ortogonal vektor}\: \vec{p}\\ &\textrm{pada}\: \: \vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \frac{3}{5}\\\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{7}{5}\\\\ &\textrm{c}.\quad \displaystyle \frac{8}{5}\\\\ &\textrm{d}.\quad \displaystyle \frac{9}{5}\\\\ &\textrm{e}.\quad \displaystyle 2\\\\ &\textrm{Jawab}\\ &\begin{aligned}\left | \vec{r} \right |&=\displaystyle \frac{\vec{p}.\vec{q}}{\left | \vec{q} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -5 \end{pmatrix}.\begin{pmatrix} 4\\ 3 \end{pmatrix}}{\sqrt{4^{2}+3^{2}}}\\ &=\displaystyle \frac{8+(-15)}{\sqrt{25}}\\ &=\left | \displaystyle \frac{-7}{5} \right |\\ &=\color{red}\displaystyle \frac{7}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Panjang Proyeksi vektor}\: \: \vec{a}=\begin{pmatrix} 5\\ 1 \end{pmatrix}\\ & \textrm{pada}\: \: \vec{b}=\begin{pmatrix} 0\\ 4 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle -1\\ &\textrm{b}.\quad -\displaystyle \frac{1}{2}\\ &\textrm{c}.\quad \color{red}1\\ &\textrm{d}.\quad \displaystyle 2\\ &\textrm{e}.\quad 4\\\\ &\textrm{Jawab}\\ &\begin{aligned}\left |\vec{c} \right |&=\left |\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} 5\\ 1 \end{pmatrix}\bullet \begin{pmatrix} 0\\ -4 \end{pmatrix}}{\left | \sqrt{0^{2}+(-4)^{2}} \right |} \right |\\ &=\left |\displaystyle \frac{0-4}{4} \right |=\left |-1 \right |=\color{red}1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Proyeksi vektor ortogonal}\: \: \vec{a}=\begin{pmatrix} 2\\ -4 \end{pmatrix}\\ & \textrm{pada}\: \: \vec{b}=\begin{pmatrix} -1\\ 2 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 2\\ -1 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 2\\ -2 \end{pmatrix}\\ &\textrm{c}.\quad \color{red}\begin{pmatrix} 2\\ -4 \end{pmatrix}\\ &\textrm{d}.\quad \begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} -2\\ 4 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{c}&=\left ( \displaystyle \frac{\vec{a}\bullet \vec{b}}{\left |\vec{b} \right |^{2}} \right ).\vec{b}\\ &=\left (\displaystyle \frac{\begin{pmatrix} 2\\ -4 \end{pmatrix}\bullet \begin{pmatrix} -1\\ 2 \end{pmatrix}}{(-1)^{2}+2^{2}} \right ).\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=\left (\displaystyle \frac{-2-8}{1+4} \right ).\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=-2\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 2\\ -4 \end{pmatrix} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
Kusnandar, Muharman, I., Indrianti, M. 2017. Pendalaman Buku Teks Matematika SMA Kelas X Peminatan MIPA. Jakarta: YUDHISTIRA.
Miyanto, Aksin, N., Suparno. 2021. Buku Interaktif Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas X Semester 2. Yogyakarta: INTAN PARIWARA.
Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
Yuana, R.A., Indriyastuti. 2017. Persektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI MANDIRI.

Contoh 6 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 26.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} -2\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} 8\\ 4 \end{pmatrix},\: \: \textrm{maka}\\ &\textrm{sudut yang dibentuk vektor}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\\ &\textrm{adalah}....\\ &\textrm{a}.\quad 0^{\circ}\\ &\textrm{b}.\quad 60^{\circ}\\ &\textrm{c}.\quad 45^{\circ}\\ &\textrm{d}.\quad 60^{\circ}\\ &\textrm{e}.\quad \color{red}90^{\circ}\\\\ &\textrm{Jawab}\\ & \begin{aligned}\vec{p}.\vec{q}&=\displaystyle \begin{vmatrix} \vec{p} \end{vmatrix}.\begin{vmatrix} \vec{q} \end{vmatrix}.\cos \angle \left (\vec{p},\, \vec{q} \right )\\ \cos \angle \left (\vec{p},\, \vec{q} \right )&=\displaystyle \frac{\vec{p}.\vec{q}}{\left | \vec{p} \right |.\left | \vec{q} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -2\\ 1 \end{pmatrix}.\begin{pmatrix} 8\\ 4 \end{pmatrix}}{\sqrt{(-2)^{2}+1^{2}}.\sqrt{8^{2}+4^{2}}}\\ &=\displaystyle \frac{-16+16}{\sqrt{20}.\sqrt{80}}\\ &=\displaystyle \frac{0}{40}\\ &=0\\ \cos \angle \left (\vec{p},\, \vec{q} \right )&=\cos 90^{\circ}\\ \angle \left (\vec{p},\, \vec{q} \right )&=\color{red}90^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Jika}\: \: \overline{OA}=\begin{pmatrix} 1\\ 2 \end{pmatrix},\: \overline{OB}=\begin{pmatrix} 4\\ 2 \end{pmatrix},\: \textrm{dan}\\ &\theta =\angle \left ( \overline{OA},\: \overline{OB} \right ),\: \textrm{maka}\: \: \tan \theta =....\\\\ &\textrm{a}.\quad \displaystyle \frac{3}{5}\\\\ &\textrm{b}.\quad \displaystyle \frac{9}{16}\\\\ &\textrm{c}.\quad \color{red}\displaystyle \frac{3}{4}\\\\ &\textrm{d}.\quad \displaystyle \frac{4}{3}\\\\ &\textrm{e}.\quad \displaystyle \frac{16}{9}\\\\ &\textrm{Jawab}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\cos \theta &=\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 1\\ 2 \end{pmatrix}.\begin{pmatrix} 4\\ 2 \end{pmatrix}}{\sqrt{1^{2}+2^{2}}\sqrt{4^{2}+2^{2}}}\\ &=\displaystyle \frac{4+4}{\sqrt{5}.\sqrt{20}}\\ &=\displaystyle \frac{8}{10} \end{aligned}&\begin{aligned}\sin \theta &=\sqrt{1-\cos ^{2}\theta }\\ &=\sqrt{1-\left ( \displaystyle \frac{8}{10} \right )^{2}}\\ &=\sqrt{\displaystyle \frac{36}{100}}\\ &=\displaystyle \frac{6}{10}\\ & \end{aligned}\\\hline \end{array} \\ &\textrm{Selanjutnya}\\ &\begin{aligned}\tan \theta &=\displaystyle \frac{\sin \theta }{\cos \theta }\\ &=\displaystyle \frac{\displaystyle \frac{6}{10}}{\displaystyle \frac{8}{10}}\\ &=\color{red}\displaystyle \frac{3}{4} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Jika}\: \: \vec{a},\: \vec{b}\: \: \textrm{dan}\: \: \vec{c}\: \: \textrm{adalah vektor satuan dengan}\\ & \vec{a}+\vec{b}+\vec{c}=0.\: \textrm{Nilai dari}\: \: \vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{c}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle -3\\ &\textrm{b}.\quad \color{red}\displaystyle -\frac{3}{2}\\ &\textrm{c}.\quad \displaystyle 0\\ &\textrm{d}.\quad \displaystyle \frac{3}{2}\\ &\textrm{e}.\quad \displaystyle 3\\\\ &\textrm{Jawab}\\ & \begin{aligned}\textrm{Karena}&\left\{\begin{matrix} \vec{a},\vec{b},\vec{c}\: \: \textrm{adalah vektor satuan, dan}\\ \vec{a}+\vec{b}+\vec{c}=0\qquad\qquad\qquad\qquad . \end{matrix}\right.\\ \textrm{segitig}&\textrm{a ABC adalah segitiga sama sisi}\\ \vec{a}.\vec{b}=&\left | \vec{a} \right |\left | \vec{b} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \vec{a}.\vec{c}=&\left | \vec{a} \right |\left | \vec{c} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \vec{b}.\vec{c}=&\left | \vec{b} \right |\left | \vec{c} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \textrm{Jadi, ni}&\textrm{lai dari}\: \: \vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{c}\\ &=\left ( -\frac{1}{2} \right )+\left ( -\frac{1}{2} \right )+\left ( -\frac{1}{2} \right )=\color{red}-\frac{3}{2} \end{aligned} \end{array}$

$.\: \qquad \color{blue}\textrm{berikut ilustrasinya}$

$\begin{array}{ll}\\ 29.&\textrm{Jika}\: \: \angle \left ( \vec{a},\vec{b} \right )=60^{\circ},\: \: \left |\vec{a} \right |=4\: \: \textrm{dan}\\ &\left |\vec{b} \right |=3,\: \: \textrm{maka}\: \: \vec{a}(\vec{a}-\vec{b})\: \: \textrm{adalah}....\\ &\textrm{a}.\quad 2\\ &\textrm{b}.\quad 4\\ &\textrm{c}.\quad 6\\ &\textrm{d}.\quad 8\\ &\textrm{e}.\quad \color{red}10\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{a}(\vec{a}-\vec{b})&=\vec{a}.\vec{a}-\vec{a}.\vec{b}\\ &=\left | \vec{a} \right |\left | \vec{a} \right |\cos 0^{\circ}-\left | \vec{a} \right |\left | \vec{b} \right |\cos 60^{\circ}\\ &=\left | \vec{a} \right |^{2}-\left | \vec{a} \right |\left | \vec{b} \right |.\displaystyle \frac{1}{2}\\ &=4^{2}-4.3.\displaystyle \frac{1}{2}\\ &=16-6\\ &=\color{red}10 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Tentukan}\: \: \overline{u}\bullet \overline{v}\: ,\: \textrm{jika diketahui}\\ &\textrm{a}.\quad \left |\overline{u} \right |=10,\: \left |\overline{v} \right |=8\sqrt{3},\: \: \cos \angle (\overline{v},\, \overline{u})=\displaystyle \frac{2}{5}\sqrt{3}\\ &\textrm{b}.\quad \left |\overline{u} \right |=6\sqrt{3},\: \left |\overline{v} \right |=4\sqrt{2},\: \: \cos (\overline{v},\, \overline{u})=30^{\circ}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad &\overline{u}\bullet \overline{v}=\left | \overline{u} \right |.\left | \overline{v} \right |.\cos \angle (\overline{v},\, \overline{u})\\ &\overline{u}\bullet \overline{v}=10.(8\sqrt{3}).\displaystyle \frac{2}{5}\sqrt{3}=2.8.2.3=\color{red}96\\ \textrm{b}.\quad &\overline{u}\bullet \overline{v}=\left | \overline{u} \right |.\left | \overline{v} \right |.\cos \angle (\overline{v},\, \overline{u})\\ &\overline{u}\bullet \overline{v}=(6\sqrt{3}).(4\sqrt{2}).\displaystyle \cos 30^{\circ}\\ &\qquad =(6\sqrt{3}).(4\sqrt{2}).\displaystyle \frac{1}{2}\sqrt{3}\\ &\qquad=\displaystyle \frac{6.4.3.\sqrt{2}}{2} =\color{red}36\sqrt{2}\\ \end{aligned} \end{array}$.

Contoh 5 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 21.&\textrm{Jika}\: \: \vec{g}=\begin{pmatrix} 3^{x+y}\\ 5 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{h}=\begin{pmatrix} 81\\ \displaystyle \frac{y+7}{2} \end{pmatrix}\\ & \textrm{sehingga}\: \: \vec{g}=\vec{h}\: \: \: \textrm{nilai dari}\: \: 4x-3y=.... \\ &\textrm{a}.\quad \color{red}-5\\ &\textrm{b}.\quad -1\\ &\textrm{c}.\quad 0\\ &\textrm{d}.\quad 5\\ &\textrm{e}.\quad 10\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{g}=\vec{h}\\ \begin{pmatrix} 3^{x+y}\\ 5 \end{pmatrix}&=\begin{pmatrix} 81\\ \displaystyle \frac{y+7}{2} \end{pmatrix}\\ 3^{x+y}&=81=3^{4}\Leftrightarrow x+y=4\\ \displaystyle \frac{y+7}{2}&=5\Leftrightarrow y=10-7=3,\quad \textrm{sehingga}\\ x+y&=4\Leftrightarrow x+3=4\Leftrightarrow x=4-3=1,\\ &\textrm{maka}\\ 4x-3y&=4(1)-3(3)\\ &=4-9=\color{red}-5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Vektor}\: \: \vec{m}=\begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{searah }\\ &\textrm{dengan vektor}\: .... \\ &\textrm{a}.\quad \displaystyle \begin{pmatrix} 2\\ -5 \end{pmatrix}\\ &\textrm{b}.\quad \displaystyle \begin{pmatrix} 2\\ 5 \end{pmatrix}\\ &\textrm{c}.\quad \color{red}\displaystyle \begin{pmatrix} -6\\ 15 \end{pmatrix}\\ &\textrm{d}.\quad \displaystyle \begin{pmatrix} -4\\ 5 \end{pmatrix}\\ &\textrm{e}.\quad \displaystyle \begin{pmatrix} -3\\ 10 \end{pmatrix} \\\\ &\textrm{Jawab}\\ &\begin{aligned}&\textrm{Vektor}\quad \vec{m}\: \: \: \textrm{searah dengan vektor}\: \: k.\vec{m}\\ &\color{blue}k.\vec{m}=k\begin{pmatrix} -2\\ 5 \end{pmatrix}, \color{black}\textrm{dengan}\: \: k\: \: \textrm{skalar positif}\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}\\\hline \begin{pmatrix} 2\\ -5 \end{pmatrix}=-\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} 2\\ 5 \end{pmatrix}=...\\\hline \textrm{c}&\textrm{d}\\\hline \color{red}\begin{pmatrix} -6\\ 15 \end{pmatrix}=3\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} -4\\ 5 \end{pmatrix}=...\\\hline \textrm{e}&\\\hline \begin{pmatrix} -3\\ 10 \end{pmatrix}=...&\\\hline \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Jika vektor}\: \: \overline{AC}=\vec{p},\: \overline{BC}=\vec{q}\: \: \textrm{dan}\\ &\overline{AD}:\overline{DC}=1:2,\: \: \textrm{maka vektor}\\ & \overline{BD}\: \: \textrm{bila dinyatakan}\\ &\textrm{dalam}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{3}\left ( 3\vec{p}-2\vec{q} \right )\\ \textrm{b}.\quad \color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right )\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-2\vec{q} \right )\\ \textrm{d}.\quad \displaystyle \left ( \vec{p}-\displaystyle \frac{1}{3}\vec{q} \right )\\ \textrm{e}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-\vec{q} \right ) \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\overline{AC}:\overline{CD}&=3:-2\\ \overline{CD}&=-\displaystyle \frac{2}{3}\, \overline{AC}\\ &=\displaystyle -\frac{2}{3}\vec{p}\\ \textrm{maka}\, ,\qquad&\\ \overline{BD}&=\overline{BC}+\overline{CD}\\ &=\vec{q}+\left ( -\displaystyle \frac{2}{3}\vec{p} \right )\\ &=\color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right ) \end{aligned} \end{array}$

$.\: \quad \color{blue}\textrm{Gambar berikut untuk soal 24}$

$\begin{array}{ll}\\ 24.&\textrm{Jika vektor}\: \: \overline{AC}=\vec{p},\: \overline{BC}=\vec{q}\: \: \textrm{dan}\\ & \overline{AD}:\overline{DC}=1:2,\: \textrm{maka vektor}\: \: \overline{BD}\\ &\textrm{bila dinyatakan dalam}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\: \: \textrm{adalah}.... \\ &\textrm{a}.\quad \displaystyle \frac{1}{3}\left ( 3\vec{p}-2\vec{q} \right )\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right )\\ &\textrm{c}.\quad \displaystyle \left ( \vec{p}-\displaystyle \frac{1}{3}\vec{q} \right )\\ &\textrm{d}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-2\vec{q} \right )\\ &\textrm{e}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-\vec{q} \right )\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{AC}:\overline{CD}&=3:-2\\ \overline{CD}&=-\displaystyle \frac{2}{3}\, \overline{AC}\\ &=\displaystyle -\frac{2}{3}\vec{p}\\ \textrm{maka}\, ,\qquad&\\ \overline{BD}&=\overline{BC}+\overline{CD}\\ &=\vec{q}+\left ( -\displaystyle \frac{2}{3}\vec{p} \right )\\ &=\color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Jika titik}\: \: A(2,6)\: \: \textrm{dan}\: \: B(5,3)\: \: \textrm{demikian }\\ &\textrm{juga titik}\: \: P\: \: \textrm{terletak pada}\: \: \overline{AB}\\ &\textrm{dengan}\: \: \overline{AP}:\overline{PB}=2:1,\: \textrm{maka vektor }\\ &\textrm{posisi}\: \: \vec{p}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \color{red}\begin{pmatrix} 4\\ 4 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 4\\ 5 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} -4\\ 4 \end{pmatrix}\\ &\textrm{d}.\quad \begin{pmatrix} 4\\ 2 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} -4\\ 6 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{AP}:\overline{PB}&=2:1\\ \overline{AP}&=2\, \overline{PB}\\ \vec{p}-\vec{a}&=2\left ( \vec{b}-\vec{p} \right )\\ \vec{p}+2\vec{p}&=\vec{a}+2\vec{b}\\ 3\vec{p}&=\vec{a}+2\vec{b}\\ \vec{p}&=\displaystyle \frac{1}{3}\left ( \vec{a}+2\vec{b} \right )\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 2+2.5\\ 6+2.3 \end{pmatrix}\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 12\\ 12 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 4\\ 4 \end{pmatrix} \end{aligned} \end{array}$

Contoh 4 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 16.&\textrm{Diketahui titik A(-1,1,0) dan titik B(1,-2,2)}\\ &\textrm{maka panjang vektor}\: \: \overrightarrow{BA}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{2}&&&\textrm{d}.&\color{red}\sqrt{17}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\sqrt{9}&\textrm{e}.&\sqrt{21} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui}\: \, \textrm{sebagaimana pada soal}\\ &\textrm{maka pan}\textrm{jang vektor}\: \: \overrightarrow{BA}\: \: \textrm{adalah}\\ &\left | \overrightarrow{BA} \right |=\sqrt{(1-(-1))^{2}+(-2-1)^{2}+(2-0)^{2}}\\ &=\sqrt{2^{2}+(-3)^{2}+2^{2}}\\ &=\sqrt{4+9+4}\\ &=\color{red}\sqrt{17} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Vektor satuan untuk vektor}\: \: \vec{a}=\begin{pmatrix} 2, & 1, &-2 \end{pmatrix}=\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} -\displaystyle \frac{2}{3}, & -\displaystyle \frac{1}{3}, &\displaystyle \frac{2}{3} \end{pmatrix}&\\\\ \textrm{b}.&\color{red}\begin{pmatrix} \displaystyle \frac{2}{3}, & \displaystyle \frac{1}{3}, &-\displaystyle \frac{2}{3} \end{pmatrix}&\\\\ \textrm{c}.&\begin{pmatrix} \displaystyle \frac{2}{4}, & \displaystyle \frac{1}{4}, &-\displaystyle \frac{2}{4} \end{pmatrix}&\\\\ \textrm{d}.&\begin{pmatrix} -\displaystyle \frac{2}{4}, & -\displaystyle \frac{1}{4}, &\displaystyle \frac{2}{4} \end{pmatrix}\\\\ \textrm{e}.&\begin{pmatrix} -\displaystyle \frac{2}{9}, &-\displaystyle \frac{1}{9}, & \displaystyle \frac{2}{9} \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor }&\: \textrm{satuan dari vektor}\: \: \vec{a}\: \: \textrm{adalah}:\\ \hat{a}&=\displaystyle \frac{\vec{a}}{\left | \vec{a} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{\sqrt{2^{2}+1^{2}+(-2)^{2}}}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{\sqrt{9}}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{3}\\ &=\color{red}\begin{pmatrix} \displaystyle \frac{2}{3}, & \displaystyle \frac{1}{3}, &-\displaystyle \frac{2}{3} \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Jika titik A(-2,3,5) dan B(4,1,-3)},\\ & \textrm{maka vektor posisi AB adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} -6, & 2, &8 \end{pmatrix}&\\ \textrm{b}.&\begin{pmatrix} 8, & 2, &-6 \end{pmatrix}&\\ \textrm{c}.&\color{red}\begin{pmatrix} 6, & -2, &-8 \end{pmatrix}&\\ \textrm{d}.&\begin{pmatrix} -8 & -2, &6 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 2, &4, & 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor posisi}&\: \textrm{dari}\: \: \overrightarrow{AB}\: \: \textrm{adalah}:\\ \overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\ &=\begin{pmatrix} 4\\ 1\\ -3 \end{pmatrix}-\begin{pmatrix} -2\\ 3\\ 5 \end{pmatrix}\\ &=\begin{pmatrix} 4+2\\ 1-3\\ -3-5 \end{pmatrix}\\ &=\begin{pmatrix} 6\\ -2\\ -8 \end{pmatrix}\quad\: \textbf{atau}\\ &=\color{red}\begin{pmatrix} 6, & -2, & -8 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} ^{2}\log 8x\\ \left ( ^{2}\log x \right )^{y} \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} 5\\ 8 \end{pmatrix}\\ &\textrm{sehingga}\: \: \vec{p}=\vec{q}\: \: \: \textrm{nilai dari}\: \: x.y=.... \\ &\textrm{a}.\quad 6\\ &\textrm{b}.\quad \color{red}12\\ &\textrm{c}.\quad 18\\ &\textrm{d}.\quad 24\\ &\textrm{e}.\quad 30\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{p}=\vec{q}\\ \begin{pmatrix} ^{2}\log 8x\\ \left ( ^{2}\log x \right )^{y} \end{pmatrix}&=\begin{pmatrix} 5\\ 8 \end{pmatrix},\quad\: \: \textrm{maka}\\ 8x&=2^{5}=32\\ \Leftrightarrow x&=\displaystyle \frac{32}{8}=4\\ \left (^{2}\log 4 \right )^{y}&=8\\ \Leftrightarrow 2^{y}&=8=2^{3}\\ \Leftrightarrow y&=3\\ \textrm{Sehingga}&\\ x.y&=4\times 3=\color{red}12 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} 3x\\ 4x+y \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} \displaystyle \frac{2x-4}{2}\\ 6 \end{pmatrix}\\ & \textrm{sehingga}\: \: \vec{p}=\vec{q}\: \: \: \textrm{nilai dari}\: \: 2x+y=.... \\ &\textrm{a}.\quad -12\\ &\textrm{b}.\quad 0\\ &\textrm{c}.\quad 8\\ &\textrm{d}.\quad \color{red}9\\ &\textrm{e}.\quad 19\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{p}=\vec{q}\\ \begin{pmatrix} 3x\\ 4x+y \end{pmatrix}&=\begin{pmatrix} \displaystyle \frac{2x-4}{2}\\ 6 \end{pmatrix}\\ 3x&=\displaystyle \frac{2x-4}{2}\Leftrightarrow 6x=2x-4\\ \Leftrightarrow x&=-1\\ 4(-1)+y&=6\Leftrightarrow -4+y=6\\ \Leftrightarrow y&=6+4\\ y&=10\\ x+y&=(-1)+10=\color{red}9 \end{aligned} \end{array}$

Contoh 3 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 11.&\textrm{Jika vektor}\: \: \vec{a}=\begin{pmatrix} 6\\ -4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 3\\ 2 \end{pmatrix},\\ &\textrm{maka}\: \: 3\vec{a}-2\vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\begin{pmatrix} 12\\ -16 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} 24\\ 16 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 24\\ -16 \end{pmatrix}&\textrm{c}.\quad \begin{pmatrix} 12\\ 16 \end{pmatrix}&\textrm{e}.\quad \begin{pmatrix} -12\\ -16 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}3\vec{a}-2\vec{b}&=3\begin{pmatrix} 6\\ -4 \end{pmatrix}-2\begin{pmatrix} 3\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 18-6\\ -12-4 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 12\\ -16 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Diketahui jajar genjang ABCD }\\ &\textrm{dengan titik E adalah perpotongan }\\ &\textrm{diagonal jajar genjang}. \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{Jika}\: \: \overline{AB}=\vec{b}\: \: \textrm{dan}\: \: \overline{AD}=\vec{a},\: \textrm{maka}\: \: \overline{CE}\\ & \textrm{bila dinyatakan dalam}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right )&\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}-\vec{b} \right )&\\ \textrm{c}.\quad \displaystyle \frac{1}{2}\left ( \vec{b}-\vec{a} \right )&\\ \textrm{d}.\quad \color{red}\displaystyle -\frac{1}{2}\left ( \vec{a}+\vec{b} \right )\\ \textrm{e}.\quad -\displaystyle \frac{1}{2}\left ( 2\vec{a}+\vec{b} \right ) \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned} \overline{AC}&=\overline{AD}+\overline{DC}\\ \overline{CA}&=\overline{CD}+\overline{DA}\\ \overline{CE}&=\displaystyle \frac{1}{2}\, \overline{CA}\\ &=\displaystyle \frac{1}{2}\left ( -\vec{b}-\vec{a} \right )\\ &=\color{red}-\displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Pada segi enam beraturan ABCDEF},\\ & \textrm{jika}\: \: \overrightarrow{AB}=\vec{u}\: \: \textrm{dan}\: \: \overrightarrow{AF}=\vec{v}\: \: \textrm{maka vektor}\\ &\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&2\vec{u}+2\vec{v}&&&\textrm{d}.&\color{red}6\vec{u}+6\vec{v}\\ \textrm{b}.&4\vec{u}+4\vec{v}&\textrm{c}.&5\vec{u}+5\vec{v}&\textrm{e}.&8\vec{u}+8\vec{v} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$

$\begin{aligned}.\: \, \qquad &\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}\\ &=\overrightarrow{AB}+\left (\overrightarrow{AO}+\overrightarrow{OC} \right )+2\overrightarrow{AO}+\left (\overrightarrow{AO}+\overrightarrow{OE} \right )+\overrightarrow{AF}\\ &=\vec{u}+\left (2\vec{u}+\vec{v} \right )+2\left ( \vec{v}+\vec{u} \right )+\left ( 2\vec{v}+\vec{u} \right )+\vec{v}\\ &=\color{red}6\vec{u}+6\vec{v} \end{aligned}$

$\begin{array}{ll}\\ 14.&\textrm{Perhatikanlah juga ilustrasi gambar berikut} \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{maka vektor}\: \: \vec{w}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\color{red}8\vec{i}-6\vec{j}-13\vec{k}&\\ \textrm{b}.&8\vec{i}-13\vec{j}-6\vec{k}\\ \textrm{c}.&6\vec{i}-8\vec{j}-13\vec{k}&\\ \textrm{d}.&-6\vec{i}+8\vec{j}-13\vec{k}\\ \textrm{e}.&-6\vec{i}-13\vec{j}+8\vec{k} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Kita perhatikan juga ilustrasi }\\ &\textrm{gambarnya semisal dengan soal No.1}\\ &\textrm{Misalkan titiknya adalah titik }\\ &\textrm{W dengan koordinat (8,-6,-13)},\\ &\textrm{maka vektor posisi titik }\\ &\textrm{W tersebut adalah}\: \: \overrightarrow{OW}=\vec{w}\\ & \textrm{di mana}\\ &\begin{aligned}&\textrm{Vektor}\: \: \vec{w}\: \textrm{jika dinyatakan }\\ &\textrm{dalam kombinasi linear adalah}\\ &\vec{w}=\color{red}8\vec{i}-6\vec{j}-13\vec{k} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Jika titik Z(4,-5,2)},\: \textrm{maka panjang }\\ &\textrm{vektor posisi titik Z adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1&&&\textrm{d}.&5\sqrt{2}\\ \textrm{b}.&2\sqrt{5}&\textrm{c}.&\color{red}3\sqrt{5}&\textrm{e}.&5\sqrt{3} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor posisi}&\: \, \textrm{titik Z tersebut adalah}\\ \overrightarrow{OZ}=\vec{z}&=\begin{pmatrix} 4, & -5, & 2 \end{pmatrix},\\ \textrm{Dan panjang}&\: \, \textrm{vektor}\: \: \vec{z}\: \: \textrm{ini adalah}\\ \left | \vec{z} \right |&=\sqrt{4^{2}+(-5)^{2}+2^{2}}\\ &=\sqrt{16+25+4}\\ &=\sqrt{45}=\sqrt{9\times 5}\\ &=\color{red}3\sqrt{5} \end{aligned} \end{array}$.

Contoh Soal 16 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 76.&\textbf{(IMO 1983)}\\ &\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah panjang sisi-sisi segitiga}\\ &\textrm{tunjukkan bahwa}\\ &\quad a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\geq 0\\\\ &\textbf{Bukti}:\\ &\textrm{Pada sebuah segitiga dengan sisi}\: \: a,b,c\\ &\textrm{berlaku}\: \: \begin{cases} a+b>c & \Rightarrow a>c-b\\ a+c>b & \Rightarrow c>b-c\\ b+c>a & \Rightarrow b>a-c \end{cases}\\ &\textrm{Sehingga untuk ketaksamaan pada soal}\\ &\color{red}a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(x-a)\\ &\geq a^{2}(a-c)(a-b)+b^{2}(b-c)(b-c)+c^{2}(c-b)(c-a)\color{red}\geq 0\\ &\textrm{Bentuk terakhir memenuhi bentuk dari}\\ &\textbf{Ketaksamaan Schur}\: \: \textrm{saat}\: \: \color{red}r=2.\\ &\textrm{Jadi},\\ &\quad a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\geq 0\quad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 77.&\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real positif dengan}\\ &a+b+c=2\: ,\: \textrm{tunjukkan bahwa}\\ &\quad a^{4}+b^{4}+c^{4}+abc\geq a^{3}+b^{3}+c^{3}\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=2\\ &\textrm{kita memiliki}\\ &a^{2}(a-b)(a-c)+b^{2}(b-a)(b-c)+c^{2}(c-a)(c-b)\geq 0\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc(a+b+c)-a^{3}(b+c)-b^{3}(a+c)-c^{3}(a+b)\geq 0\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc(a+b+c)\geq a^{3}(b+c)+b^{3}(a+c)+c^{3}(a+b)\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc(a+b+c)\geq (a^{3}+b^{3}+c^{3})(a+b+c)-(a^{4}+b^{4}+c^{4})\\ &\Leftrightarrow 2(a^{4}+b^{4}+c^{4})+abc(a+b+c)\geq (a^{3}+b^{3}+c^{3})(a+b+c)\\ &\Leftrightarrow 2(a^{4}+b^{4}+c^{4})+abc(2)\geq (a^{3}+b^{3}+c^{3})(2)\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc\geq a^{3}+b^{3}+c^{3}\qquad \blacksquare \\\\ &\color{blue}\textrm{Bentuk di atas kadang dituliskan dengan bentuk}\\ &\color{blue}\textrm{berikut}:\\ &\begin{aligned}&\textrm{Dengan}\: \: \textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=2\\ &\textrm{kita memiliki}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{2}(a-b)(a-c)\geq 0\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a-\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}(b+c)\geq 0\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a\geq \displaystyle \sum_{\textrm{siklik}}^{.}a^{3}(b+c)\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{3} \right )\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )-\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{4} \right )\\ &\Leftrightarrow 2\displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{3} \right )\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{3} \right )\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 78.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad a^{2}+b^{2}+c^{2}+2abc+1\geq 2(ab+acb+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Darij Grinberg})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{dilanjutkan dengan}\: \: \textbf{ketaksamaan Schur}\\ &\textrm{serta menggesernya ke ruas kiri, maka}\\ &a^{2}+b^{2}+c^{2}+2abc+1- 2(ab+ac+bc)\\ &\geq a^{2}+b^{2}+c^{2}+3(abc)^{^{\frac{2}{3}}}+1\geq 2(ab+ac+bc)\\ &\geq \left ((a)^{^{\frac{2}{3}}} \right )^{3}+\left ((b)^{^{\frac{2}{3}}} \right )^{3}+\left ((c)^{^{\frac{2}{3}}} \right )^{3}+3(abc)^{^{\frac{2}{3}}}-2(ab+ac+bc)\\ &\geq a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +b^{.^{\frac{2}{3}}} \right )+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )-2(ab+ac+bc)\\ &\geq a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +b^{.^{\frac{2}{3}}} \right )+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )-2(ab+ac+bc)\\ &= a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} -b^{.^{\frac{2}{3}}} \right )^{2}+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} -c^{.^{\frac{2}{3}}} \right )^{2}\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} -c^{.^{\frac{2}{3}}} \right )^{2}\geq 0\qquad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 79.&(\textbf{APMO 2004})\\ &\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad (x^{2}+2)(y^{2}+2)(z^{2}+2)\geq 9(xy+yz+zx)\\\\ &\textbf{Bukti}\\ &\color{blue}\textbf{Alternatif 1}\\ &\textrm{Dengan menjabarkan akan didapatkan}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{Perhatikan bahwa}\\ &\bullet \quad\color{red}2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}-4\displaystyle \sum_{\textrm{siklik}}^{.}xy+6=2\displaystyle \sum_{\textrm{siklik}}^{.}(xy-1)^{2}\geq 0\\ &\bullet \quad \color{red}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\: \: \color{black}\textrm{atau}\: \: \color{red}x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\textrm{Kita cukup membuktikan bahwa}\\ & x^{2}y^{2}z^{2}+\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+2\geq 2\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\Leftrightarrow x^{2}y^{2}z^{2}+2\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\begin{aligned}&\textrm{Untuk}\: \: a,b,c\: \: \textrm{bilangan real positif},\\ &\textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=1\\ &\textrm{memberikan}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}+3abc\geq \displaystyle \sum_{\textrm{siklik}}^{.}a^{2}b+\displaystyle \sum_{\textrm{siklik}}^{.}ab^{2}\\ &\qquad\qquad\qquad =ab(a+b)+bc(b+c)+ca(c+a)\\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan AM-GM}\: \: \textrm{didapakan}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}+3abc\geq 2\displaystyle \sum_{\textrm{siklik}}^{.}(ab)^{\frac{3}{2}}\\ &\textrm{Pilih}\: \: a=x^{\frac{2}{3}},\: b=y^{\frac{2}{3}},\: c=z^{\frac{2}{3}},\: \textrm{maka didapatkan}\\ &(x^{\frac{2}{3}})^{3}+(y^{\frac{2}{3}})^{3}+(z^{\frac{2}{3}})^{3}+3(xyz)^{\frac{2}{3}}\geq 2(xy+yz+zx)\\ &\textrm{Selanjutnya kita selesaikan ini}\: ,\: x^{2}y^{2}z^{2}+2\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\Leftrightarrow x^{2}y^{2}z^{2}+2 \geq 3(xyz)^{\frac{2}{3}}\\ &\textrm{Misalkan}\: \: (xyz)^{\frac{2}{3}}=t,\: \textrm{maka}\\ &t^{3}+2\geq 3t\Leftrightarrow t^{3}-3t+2\geq 0\\ &(t-1)^{2}(t+2)\geq 0\: \: \textrm{adalah hal benar} \end{aligned} \end{array}$.

$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{atau dalam bentuk utuhnya, yaitu}\\ &x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\geq 9(xy+xz+yz)\\ &\textrm{Sekarang kira uraikan satu persatu bagian}\\ &\bullet\quad x^{2}y^{2}z^{2}+1+1\geq 3\sqrt[3]{(xyz)^{2}}\geq \displaystyle \frac{9abc}{a+b+c}=\frac{9r}{p}\\ &\qquad \textrm{ingat bahwa}\: \: \textbf{jika ada}\: \: \displaystyle \frac{9r}{p}\geq 4q-p^{2}\\ &\qquad =4(xy+xz+yz)-(x+y+z)^{2}\\ &\qquad \textrm{adalah}\: \: \textbf{ketaksamaan Schur saat}\: \: \color{red}r=1\\ &\bullet \quad x^{2}y^{2}+1+x^{2}z^{2}+1+y^{2}z^{2}+1\geq 2(xy+xz+yz)\\ &\bullet\quad x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\qquad \textrm{keduanya didapat dengan ketaksamaan}\: \: \textbf{AM-GM}\\&x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\\ &=x^{2}y^{2}z^{2}+2+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}+3)+4(x^{2}+y^{2}+z^{2})\\ &\geq 4(xy+xz+yz)-(x+y+z)^{2}+4(xy+xz+yz)+4(xy+xz+yz)\\ &\geq 12(xy+xz+yz)-(x+y+z)^{2}\\ &\geq 12(xy+xz+yz)-3(xy+xz+yz)\\ &=9(xy+xz+yz)\qquad \blacksquare \end{aligned}$.

$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 3}\\ &(x^{2}+2)(y^{2}+2)(z^{2}+2)- 9(xy+yz+zx)\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\\ &=x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})\\ &\quad +4(x^{2}+y^{2}+z^{2})+8- 9(xy+xz+yz)\\ &=4(x^{2}+y^{2}+z^{2})+2\left ((x^{2}y^{2}+1) +(x^{2}z^{2}+1)+(y^{2}z^{2}+1) \right )\\ &\quad +(x^{2}y^{2}z^{2}+1)+1-9(xy+xz+yz)\\ &\geq 4(x^{2}+y^{2}+z^{2})+4(xy+xz+yz)\\ &\quad +2xyz+1-9(xy+xz+yz)\\ &=(x^{2}+y^{2}+z^{2})+3(x^{2}+y^{2}+z^{2})\\ &\quad +2xyz+1-5(x^{2}+y^{2}+z^{2})\\ &\geq x^{2}+y^{2}+z^{2}+3(xy+xz+yz)\\ &+2xyz-5(xy+xz+yz)\\ &=x^{2}+y^{2}+z^{2}+2xyz+1-2(xy+xz+yz)\geq 0\\ &\textrm{adalah benar dengan bukti ada pada}\\ &\textrm{nomor soal sebelumnya} \end{aligned}$.

$\begin{array}{ll}\\ 80.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad 2(a^{2}+b^{2}+c^{2})+abc+8\geq 5(a+b+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Tran Nam Dung})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{menggeser ke ruas kiri dan masing-masing}\\ &\textrm{serta mengalikan semunya dengan 6, maka}\\ &12(a^{2}+b^{2}+c^{2})+6abc+48- 30(a+b+c)\\ &= 12(a^{2}+b^{2}+c^{2})+3(2abc+1)+45- 5.2.3(a+b+c)\\ &\geq 2(a^{2}+b^{2}+c^{2})+9\sqrt[3]{(abc)^{2}}+45- 5\left ((a+b+c)^{2}+9 \right )\\ &=12(a^{2}+b^{2}+c^{2})+\displaystyle \frac{9abc}{\sqrt[3]{abc}}-5\left ((a^{2}+b^{2}+c^{2})+2(ab+ac+bc) \right )\\ &=7(a^{2}+b^{2}+c^{2})+ \displaystyle \frac{9abc}{\sqrt[3]{abc}}-10(ab+ac+bc)\\ &\geq 7(a^{2}+b^{2}+c^{2})+\displaystyle \frac{27abc}{a+b+c}-10(ab+ac+bc)\\ &\begin{aligned}&\textrm{dengan}\: \: \textbf{ketaksamaan Schur},\: \: \textrm{yaitu}:\\ &p^{3}+9r\geq 4pq\Leftrightarrow \displaystyle \color{red}\frac{9r}{p}\geq 4q-p^{2}\\ &\textrm{maka ketaksamaan akan menjadi}\\ &\geq 7(a^{2}+b^{2}+c^{2})+\color{blue}3(4q-p^{2})-10q\\ &\geq 7(a^{2}+b^{2}+c^{2})+2q-3p^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2(ab+ac+bc)-3(a+b+c)^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2q-3\left ((a^{2}+b^{2}+c^{2})+2q \right )\\ &=4(a^{2}+b^{2}+c^{2})+2q-6q\\ &=4(a^{2}+b^{2}+c^{2})-4q\\ &=4(a^{2}+b^{2}+c^{2})-4(ab+ac+bc)\\ &=4(a^{2}+b^{2}+c^{2}-ab-ac-bc)\geq 0\quad \blacksquare \end{aligned} \end{array}$.

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