Lanjutan Materi Polinom

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Polinom

A. Bentuk Umum

Perhatikanlah bentuk berkut

adalah bentuk suku banyak dalam variabel x berderajat n (berpangkat tertinggi).

Selanjutnya perhatikanlah tabel berikut:

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{f(x)=\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0}}\\\hline \begin{aligned}a_{n}&\: \: \textrm{adalah koefisien dari} \: \: x^{n}\\ a_{n-1}&\: \: \textrm{adalah koefisien dari} \: \: x^{n-1}\\ a_{n-2}&\: \: \textrm{adalah koefisien dari} \: \: x^{n-2}\\ \vdots &\\ a_{2}&\: \: \textrm{adalah koefisien dari} \: \: x^{2}\\ a_{1}&\: \: \textrm{adalah koefisien dari} \: \: x^{1}\\ a_{0}&\: \: \textrm{adalah konstanta(suku tetap)} \\\end{aligned}&\begin{aligned}a_{n}\: &\: \neq 0\\ n:&\: \: \textrm{bilangan cacah},\\ :&\: \: \textrm{adalah derajat (pangkat)} \\ &\: \: \textrm{tertinggi dalam suku banyak} \\ &\: \: \textrm{tersebut}\\ &\\ &\end{aligned}\\\hline \multicolumn{2}{|l|}{\begin{aligned}&\\ \textrm{Nilai}\: &\: \textrm{suku banyak}\: \: f(x)\: \: \textrm{berderajat n saat}\: \: x = k\: \: \textrm{adalah}\: \: f(k).\\ &\textrm{Jika}\: \: f(k)=0\: \: \textrm{maka}\: \: x = k\: \: \textrm{akar dari}\: \: f(x),\\ &\textrm{dan}\: \: (x-k)\: \: \textrm{faktor dari}\: \: f(x)\\ &\end{aligned}}\\\hline \end{array}.

B. Pembagian dan Nilai Suku Banyak

Ada 2 cara, yaitu:

  1. Pembagian biasa (cara bersusun atau bentuk panjang)
  2. Metode pembagian sintetis atau pembagian cara Horner dan atau Horner-Kino

Perhatikanlah ilustrasi berikut:

Berikut contoh pembagian cara biasa,

Jika contoh pembagian cara biasa di atas kita tampilkan dengan model Horner, maka akan berupa ilustrasi sebagai berikut:

Perlu diingat di sini bahwa jika f(x) adalah suku banyak berderajat n dan p(x) suku banyak berderajat k dengan  k < n , maka akan menghasilkan hasil bagi h(x) dengan sisa s(x),

f(x) = p(x) . h(x) + s(x)

dengan derajat h(x) adalah (n – k) dan derajat s(x) kurang dari k.

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Lanjutan 2 Contoh Soal Mid Semester Genap Kelas XI Materi Lingkaran

\begin{array}{ll}\\ 3.&\textrm{Himpunan pasangan berurutan yang ditunjukkan oleh fungsi}\\ &f:x\rightarrow 2-(x+1)^{2} \: \: \textrm{dari domain}\: \: \left \{ -1,0,1,2 \right \}\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \left \{ (-1,2),(0,3),(1,5),(2,7) \right \}\\ &\textrm{B}.\quad \left \{ (-1,2),(0,1),(1,-2),(2,-7) \right \}\\ &\textrm{C}.\quad \left \{ (-1,1),(0,-1),(1,-4),(2,7) \right \}\\ &\textrm{D}.\quad \left \{ (-1,0),(0,3),(1,-2),(2,7) \right \}\\ &\textrm{E}.\quad \left \{ (-1,0),(0,-4),(1,5),(2,-7) \right \}\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ &\begin{aligned}f:x\rightarrow &2-(x+1)^{2},\quad \textrm{ini artinya}\: \: f(x)=2-(x+1)^{2}\\ \textrm{untuk}&\\ f(x)&=2-(x+1)^{2}\\ \textrm{maka}&\\ f(-1)&=2-(-1+1)^{2}=2-0=2\\ f(0)&=2-(0+1)^{2}=2-1=1\\ f(1)&=2-(1+1)^{2}=2-4=-2\\ f(2)&=2-(2+1)^{2}=2-9=-7 \end{aligned} \end{array}
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Lanjutan Contoh Soal Mid Semester Genap Kelas XI 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 b^{2}\\\\ &\textrm{Jawab}:\qquad \textbf{E}\\ &\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\\ (2b)^{2}-4.1.c&=0\\ 4c&=4b^{2}\\ c&=b^{2} \end{aligned}\\\\ &\textbf{Atau sebagai alternatif jawaban yang lain}\\\\ &\begin{aligned}x^{2}+y^{2}+2ax+2by+c\: \: &=0\\ x^{2}+2ax+a^{2}+y^{2}+2by&+b^{2}+c-a^{2}-b^{2}=0\\ (x+a)^{2}+(y+b)^{2}&=a^{2}+b^{2}-c\\ \textrm{Karena menyinggung su}&\textrm{mbu-Y, maka}\: \: R=a \\ \textrm{Sehingga}\: \: R^{2}=a^{2}+b^{2}-&c=a^{2}\\ b^{2}-c&=0\\ b^{2}&=c \end{aligned} \end{array}.

\begin{array}{ll}\\ 12.&\textrm{Diketahui pusat lingkaran L terletak dikuadran I dan berada di sepanjang garis}\: \: y=2x.\\ &\textrm{Jika 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 x^{2}+y^{2}-6x-12y+36=0\\\\ &\textrm{Jawab}:\qquad \textbf{E}\\ &\begin{aligned}(x-a)^{2}+(y-b)^{2}=r^{2},\quad \: \, \, &\\ \textrm{menyinggung titik}\: \: (0,6)\Rightarrow \, &\textrm{berarti pusat lingkaran L juga terletak pada garis}\: \: y=6.\\ \textrm{ini berarti pusat lingkaran}\: \: \, &\: \textrm{L berpusat di}\: \: (x,2x)=(\frac{y}{2},y),\: \: \textrm{dengan}\\ y=6.\, \: \textrm{Sehingga pusatnya b}&\textrm{erada pada titik}\: \: (3,6).\\ \textrm{Maka persamaan lingkaran}\: \, &\textrm{adalah}\: \: (x-3)^{2}+(y-6)^{2}=3^{2}\: \: \textrm{ingat}\: \: r=\textrm{absis}\: \: x=3\\ (x-3)^{2}+(y-6)^{2}&=x^{2}-6x+9+y^{2}+12x+36=9\\ &\Leftrightarrow \, x^{2}+y^{2}-6x+12y+36=0 \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 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\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ &\begin{aligned}x^{2}+y^{2}+8x-3y-24=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 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},\qquad \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}\\ 6(x+4)+\frac{5}{2}(y-\frac{3}{2})&=\frac{169}{4}\\ 24(x+4)+5(2y-3)&=169\\ 24x+96+10y-15&=169\\ 24x+10y&=169-96+15=88\\ 12x+5y-44&=0 \end{aligned} \end{array}.

\begin{array}{ll}\\ 14.&\textrm{Sebuah garis singgung}\: \: g\: \: \textrm{menyinggung lingkaran yang berpusat di}\: \: (-2,5)\: \: \textrm{dan}\\ &\textrm{berjari-jari}\: \: 2\sqrt{10}\: \: \textrm{di titk}\: \: (4,3).\: \textrm{Maka persamaan garis singgung}\: \: g\: \: \textrm{adalah}.... \\ &\textrm{A}.\quad y=3x+9\\ &\textrm{B}.\quad y=3x-9\\ &\textrm{C}.\quad y=-3x+9\\ &\textrm{D}.\quad y=-3x-9\\ &\textrm{E}.\quad y=3x+21\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ &\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}\\ (x_{1}+2)(x+2)+(y_{1}-5)(y-5)&=40,\qquad \textrm{menyingung garis}\: \: g\: \: \textrm{di}\: (4,3)\\ (4+2)(x+2)+(3-5)(y-5)&=40\\ 6x+12-2y+10&=40\\ 6x-2y&=40-12-10\\ 3x-y&=9\\ -y&=-3x+9\\ y&=3x-9 \end{aligned} \end{array}.

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

\begin{array}{ll}\\ 16.&\textrm{Salah satu garis singgung yang bersudut}\: \: 120^{\circ}\: \: \textrm{terhadap sumbu x positif terhadap}\\ &\textrm{lingkaran dengan ujung diameter titik}\: \: (7,6)\: \textrm{dan}\: \: (1,-2)\: \: \textrm{adalah}....\\ &\textrm{A}.\quad 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\\\\ &\textrm{Jawab}:\qquad \textbf{A}\\ &\begin{array}{|c|c|c|}\hline \textrm{Pusat Lingkaran}&\textrm{Gradien Garis Singgung}&\textrm{Jari-jari}\\\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}&\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}\\\hline \multicolumn{3}{|c|}{\begin{aligned}\textrm{Sehingga persamaan }&\: \textrm{garis singgungnya adalah:}\\\\ (y-b)&=m(x-a)\pm r\sqrt{1+m^{2}}\\ (y-2)&=-\sqrt{3}(x-4)\pm 5\sqrt{1+(-\sqrt{3})^{2}}\\ y-2&=-\sqrt{3}x+4\sqrt{3}\pm 5\sqrt{1+4}\\ y&=-\sqrt{3}x+4\sqrt{3}+2\pm 10=\begin{cases} -\sqrt{3}x+4\sqrt{3}+2+ 10 \\ -\sqrt{3}x+4\sqrt{3}+2- 10 \end{cases}\\ y&=\begin{cases} -\sqrt{3}x+4\sqrt{3}+12 & \\ -\sqrt{3}x+4\sqrt{3}-8 & \end{cases} \end{aligned}}\\\hline \end{array} \end{array}.

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Contoh Soal Mid Semester Genap Kelas XI 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 4\sqrt{3}\\ &\textrm{C}.\quad 5\sqrt{2}\\ &\textrm{D}.\quad \displaystyle 6\sqrt{3}\\ &\textrm{E}.\quad 7\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ &\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 (7,-9)\\ &\textrm{D}.\quad \displaystyle (7,6)\\ &\textrm{E}.\quad (15,48)\\\\ &\textrm{Jawab}:\qquad \textbf{C}\\ &\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 (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\\\\ &\textrm{Jawab}:\: \: \textbf{A}\\ &\begin{array}{|l|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Persamaan Lingkaran Berpusat di}\: \: (a,b)\: \: \textrm{adalah}}&\\ \multicolumn{2}{|c|}{(x-a)^{2}+(y-b)^{2}=r^{2}}&\\\cline{1-2} \textrm{Pusat di}\: \: P(-2,5)&\textrm{Melalui Titik}\: \: T(3,4)&\textrm{Sehinga persamaan lingkarannya}\\\cline{1-2} \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}&\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{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 (2,-3)\: \: \textrm{dan}\: \: 3\\ &\textrm{E}.\quad (2,3)\: \: \textrm{dan}\: \: 3\\\\ &\textrm{Jawab}:\: \:\textbf{D} \\ &\textbf{Alterntif 1}\\ &\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Persamaan Lingkaran Berpusat di}\: \: (a,b)\: \: \textrm{dan berjari-jari}\: \: r\: \: \textrm{adalah}}\\ \multicolumn{2}{|c|}{\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}}\\\cline{1-2} \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 -4\\ &\textrm{E}.\quad -5\\\\ &\textrm{Jawab}:\qquad \textbf{D}\\ &\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}..

\begin{array}{ll}\\ 6.&\textrm{Diketahui lingkaran}\: \: x^{2}+y^{2}+4x+ky-12=0\: \: \textrm{melalui titik}\: \: (-2,8)\: \: \textrm{maka jari-jari}\\ &\textrm{lingkaran tersebut adalah}....\\ &\textrm{A}.\quad 1\\ &\textrm{B}.\quad 5\\ &\textrm{C}.\quad 6\\ &\textrm{D}.\quad 12\\ &\textrm{E}.\quad 25\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ &\begin{aligned}\textrm{Lingkran} \qquad \qquad\quad\quad &\\ x^{2}+y^{2}+4x+ky-12&=0\: \: \textrm{melalui}\: \: (-2,8)\: \: \textrm{berarti }\\ x^{2}+y^{2}+4x+ky-12&=(-2)^{2}+8^{2}+4(-2)+k.8-12\\ 0&=4+64-8-12+8k\\ 0&=48+8k\\ -6&=k\\ \textrm{Sehingga}\qquad\quad\quad\quad r\: \, &=\sqrt{\displaystyle \frac{4^{2}}{4}+\frac{(-6)^{2}}{4}-(-12)}=\sqrt{\displaystyle 4+9+12}=\sqrt{25}=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 adalah}....\\ &\textrm{A}.\quad (6,-4)\\ &\textrm{B}.\quad (6,6)\\ &\textrm{C}.\quad (3,-4)\\ &\textrm{D}.\quad (-6,-4)\\ &\textrm{E}.\quad (3,4)\\\\ &\textrm{Jawab}:\qquad \textbf{C}\\ &\begin{aligned}\textbf{Lingkaran}\: \: x^{2}+y^{2}+px+8y+9&=0\\ \textrm{maka,}\: \: \, \quad\quad\qquad\qquad\qquad\quad\quad\quad\: \: \: &\\ 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,}\: \: &\textrm{maka}\: \: 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=\pm 6\\ p=-6\: \Rightarrow \: x^{2}+y^{2}-6x+8y+9&=0\Rightarrow \textrm{pusatnya adalah}\: \: \left ( -\displaystyle \frac{A}{2},-\frac{B}{2} \right )=(3,-4)\\ p=6\: \: \: \, \: \Rightarrow \: x^{2}+y^{2}+6x+8y+9&=0\Rightarrow \textrm{pusatnya adalah}\: \: \left ( -\displaystyle \frac{A}{2},-\frac{B}{2} \right )=(-3,-4) \end{aligned} \end{array}.

Sebagai ilustrasi perhatikanlah gambar berikut ini

\begin{array}{ll}\\ 8.&\textrm{Titik-titik berikut yang posisinya berada di luar lingkaran}\: \: x^{2}+y^{2}-2x+8y-32=0\: \: \textrm{adalah}.... \\ &\textrm{A}.\quad (0,0)\\ &\textrm{B}.\quad (-6,-4)\\ &\textrm{C}.\quad (-3,2)\\ &\textrm{D}.\quad (3,1)\\ &\textrm{E}.\quad (4,1)\\\\ &\textrm{Jawab}:\qquad \textbf{C}\\ &\begin{array}{|c|c|l|c|}\hline \textrm{Opsi}&\textrm{Titik}&\textrm{Lingkaran}\equiv 0^{2}+0^{2}-2.0+8.0-32=-32&\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 \textcircled{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} \end{array}.

Sebagai ilustrasi perhatikanlah gambar berikut:

\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 2\sqrt{5}\\ &\textrm{C}.\quad \sqrt{10}\\ &\textrm{D}.\quad 5\\ &\textrm{E}.\quad 4\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ &\begin{aligned}\textrm{Perhatikanlah bahwa},\quad\quad\qquad\qquad\quad\quad\, &\\ x^{2}+y^{2}-4x+8y+10&=0,\quad \textrm{dengan}\: \: x=2y+5\\ (2y+5)^{2}+y^{2}-4(2y+5)+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};\qquad y&=-3\Rightarrow x=2(-3)+5=-1,\quad A(-1,-3)\\ \qquad 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}\\ &=2\sqrt{5} \end{aligned} \end{array}.

Berikut ilustrasi gambarnya:

\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 \textrm{titik pusatnya terletak pada}\: \: x-y=0\\ &\textrm{E}.\quad \textrm{berjari-jari 3}\\\\ &\textrm{Jawab}:\qquad \textbf{D}\\ &\begin{array}{|c|l|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}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} &=(3,3) \\ \textrm{Jari-jari} & =2\sqrt{3} \end{cases} \end{aligned}}\\\hline \textrm{Opsi}&\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 \textcircled{D}&\textrm{titik pusatnya terletak pada garis}\: \: x-y=0&\textbf{tepat}\\\hline \textrm{E}&\textrm{berjari-jari 3}&\textrm{tidak tepat}\\\hline \end{array} \end{array}.

Dan berikut ilustrasinya:

 

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Lanjutan 2 Contoh Soal Mid Semester Genap Kelas X Materi Vektor

\begin{array}{ll}\\ 21.&\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 \displaystyle -\frac{3}{2}\\ &\textrm{C}.\quad \displaystyle 0\\ &\textrm{D}.\quad \displaystyle \frac{3}{2}\\ &\textrm{E}.\quad \displaystyle 3\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ & \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 )=-\frac{3}{2} \end{aligned} \end{array}..

Sebagai ilustrasi perhatikanlah gambar berikut

\begin{array}{ll}\\ 22.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} 2\\ -5 \end{pmatrix},\: \vec{q}=\begin{pmatrix} 4\\ 3 \end{pmatrix},\: \textrm{maka proyeksi skalar ortogonal vektor}\: \vec{p}\: \: \textrm{pada}\: \: \vec{q}\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \displaystyle \frac{3}{5}\\ &\textrm{B}.\quad \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}:\qquad \textbf{B}\\ &\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 |\\ &=\displaystyle \frac{7}{5} \end{aligned} \end{array}.

\begin{array}{ll}\\ 23.&\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 1\\ &\textrm{D}.\quad \displaystyle 2\\ &\textrm{E}.\quad 4\\\\ &\textrm{Jawab}:\qquad \textbf{C}\\ &\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 |=1 \end{aligned} \end{array}.

\begin{array}{ll}\\ 24.&\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 \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}:\qquad \textbf{C}\\ &\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}\\ &=\begin{pmatrix} 2\\ -4 \end{pmatrix} \end{aligned} \end{array}.

 

 

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Lanjutan Contoh Soal Mid Semester Genap Kelas X Materi Vektor

\begin{array}{ll}\\ 11.&\textrm{Vektor}\: \: \vec{m}=\begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{searah 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 \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}:\qquad \textbf{C}\\ &\begin{aligned}&\textrm{Vektor}\quad \vec{m}\: \: \: \textrm{searah dengan vektor}\: \: k.\vec{m}\\\\ &\begin{array}{|c|c|c|c|c|}\hline \multicolumn{5}{|c|}{k.\vec{m}=k\begin{pmatrix} -2\\ 5 \end{pmatrix}, \textrm{dengan}\: \: k\: \: \textrm{positif}}\\\hline \textrm{a}&\textrm{b}&\textrm{c}&\textrm{d}&\textrm{e}\\\hline \begin{pmatrix} 2\\ -5 \end{pmatrix}=-\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} 2\\ 5 \end{pmatrix}=...&\begin{pmatrix} -6\\ 15 \end{pmatrix}=3\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} -4\\ 5 \end{pmatrix}=...&\begin{pmatrix} -3\\ 10 \end{pmatrix}=...\\\hline \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ 12.&\textrm{Vektor satuan}\: \: \vec{u}=\begin{pmatrix} -5\\ -12 \end{pmatrix}\: \: \textrm{adalah}.... \\ &\textrm{A}.\quad \displaystyle -\frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{B}.\quad \displaystyle \frac{1}{15}\begin{pmatrix} -5\\ -12 \end{pmatrix}\\ &\textrm{C}.\quad \displaystyle -\frac{1}{17}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{D}.\quad \displaystyle -\frac{1}{17}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{E}.\quad \displaystyle \frac{1}{2} \begin{pmatrix} 5\\ 12 \end{pmatrix}\\\\ &\textrm{Jawab}:\qquad \textbf{A}\\ &\begin{aligned}\vec{e}_{\vec{u}}&=\displaystyle \frac{\vec{u}}{\left | \vec{u} \right |},\quad \textrm{maka}\\ \vec{e}_{\begin{pmatrix} -5\\ -12 \end{pmatrix}}&=\displaystyle \frac{\begin{pmatrix} -5\\ -12 \end{pmatrix}}{\left | \begin{pmatrix} -5\\ -12 \end{pmatrix} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -5\\ -12 \end{pmatrix}}{\sqrt{(-5)^{2}+(-12)^{2}}}\\ &=\displaystyle \frac{-\begin{pmatrix} 5\\ 12 \end{pmatrix}}{\sqrt{169}}\\ &=-\displaystyle \frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix} \end{aligned} \end{array}.

\begin{array}{ll}\\ 13.&\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 12\\ &\textrm{C}.\quad 18\\ &\textrm{D}.\quad 24\\ &\textrm{E}.\quad 30\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ &\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=12 \end{aligned} \end{array}.

\begin{array}{ll}\\ 14.&\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 9\\ &\textrm{E}.\quad 19\\\\ &\textrm{Jawab}:\qquad \textbf{D}\\ &\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=9 \end{aligned} \end{array}.

\begin{array}{ll}\\ 15.&\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 -5\\ &\textrm{B}.\quad -1\\ &\textrm{C}.\quad 0\\ &\textrm{D}.\quad 5\\ &\textrm{E}.\quad 10\\\\ &\textrm{Jawab}:\qquad \textbf{A}\\ &\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,\quad \textrm{maka}\\ 4x-3y&=4(1)-3(3)\\ &=4-9=-5 \end{aligned} \end{array}.

Gambar berikut untuk soal No.16

\begin{array}{ll}\\ 16.&\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 \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}:\qquad \textbf{B}\\ &\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 )\\ &=\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right ) \end{aligned} \end{array}.

\begin{array}{ll}\\ 17.&\textrm{Jika titik}\: \: A(2,6)\: \: \textrm{dan}\: \: B(5,3)\: \: \textrm{demikian juga titik}\: \: P\: \: \textrm{terletak pada}\: \: \overline{AB}\\ &\textrm{dengan}\: \: \overline{AP}:\overline{PB}=2:1,\: \textrm{maka vektor posisi}\: \: \vec{p}\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \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}:\qquad \textbf{A}\\ &\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}\\ &=\begin{pmatrix} 4\\ 4 \end{pmatrix} \end{aligned} \end{array}.

\begin{array}{ll}\\ 18.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} -2\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} 8\\ 4 \end{pmatrix},\: \: \textrm{maka 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 90^{\circ}\\\\ &\textrm{Jawab}:\qquad \textbf{E}\\ & \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 )&=90^{\circ} \end{aligned} \end{array}.

\begin{array}{ll}\\ 19.&\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 10\\\\ &\textrm{Jawab}:\qquad \textbf{E}\\ &\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\\ &=10 \end{aligned} \end{array}.

\begin{array}{ll}\\ 20.&\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 \displaystyle \frac{3}{4}\\ &\textrm{D}.\quad \displaystyle \frac{4}{3}\\ &\textrm{E}.\quad \displaystyle \frac{16}{9}\\\\ &\textrm{Jawab}:\qquad \textbf{C}\\ &\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 \multicolumn{2}{|c|}{\begin{aligned}\tan \theta &=\displaystyle \frac{\sin \theta }{\cos \theta }\\ &=\displaystyle \frac{\frac{6}{10}}{\frac{8}{10}}\\ &=\displaystyle \frac{3}{4} \end{aligned}}\\\hline \end{array} \end{array}.

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Contoh Soal Mid Semester Genap Materi Vektor

\begin{array}{ll}\\ 1.&\textrm{Panjang vektor}\: \: \vec{p}=\begin{pmatrix} 4\\ -8 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \displaystyle \sqrt{4}\\ &\textrm{B}.\quad \sqrt{12}\\ &\textrm{C}.\quad \sqrt{20}\\ &\textrm{D}.\quad \displaystyle \sqrt{80}\\ &\textrm{E}.\quad \sqrt{100}\\\\ &\textrm{Jawab}:\qquad \textbf{D}\\ &\begin{aligned}\vec{p}&=\begin{pmatrix} 4\\ -8 \end{pmatrix},\: \: \textrm{maka besar dari vektor}\: \: \vec{p}\: \: \textrm{adalah}=\left | \vec{p} \right |=\sqrt{x^{2}+y^{2}}\\ &\textrm{Yaitu}\\ \left | \vec{p} \right |&=\sqrt{4^{2}+(-8)^{2}}\\ &=\sqrt{16+64}=\sqrt{80} \end{aligned} \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Vektor satuan dari}\: \: \vec{q}=3\vec{i}-4\vec{j}\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \displaystyle \frac{4}{5}\vec{i}-\frac{3}{5}\vec{j}\\ &\textrm{B}.\quad \displaystyle \frac{3}{5}\vec{i}-\frac{4}{5}\vec{j}\\ &\textrm{C}.\quad 3\vec{i}-4\vec{j}\\ &\textrm{D}.\quad 4\vec{i}-3\vec{j}\\ &\textrm{E}.\quad 15\vec{i}-20\vec{j}\\\\ &\textrm{Jawab}:\qquad \textbf{B}\\ &\begin{aligned}\vec{q}&=3\vec{i}-4\vec{j}\\ &\textrm{Vektor satuan dari vektor}\: \: \vec{q}\: \: \textrm{adalah}\: :\: \hat{e}_{_{\vec{q}}}=\displaystyle \frac{1}{\left | \vec{q} \right |}. \vec{q}\\ &\textrm{Sehingga}\\ \hat{e}_{_{\vec{q}}}&=\displaystyle \frac{1}{\sqrt{3^{2}+(-4)^{2}}}.\begin{pmatrix} 3\\ -4 \end{pmatrix}\\ &=\displaystyle \frac{1}{5}\begin{pmatrix} 3\\ -4 \end{pmatrix}\\ &\textrm{atau dalam vektor basis}\\ &=\displaystyle \frac{3}{5}\vec{i}-\frac{4}{5}\vec{j} \end{aligned} \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Vektor berikut yang memiliki panjang}\: \: 29\: \: \textrm{satuan adalah}....\\ &\textrm{A}.\quad \displaystyle 18\vec{i}-19\vec{j}\\ &\textrm{B}.\quad \displaystyle 19\vec{i}-20\vec{j}\\ &\textrm{C}.\quad 20\vec{i}-21\vec{j}\\ &\textrm{D}.\quad 21\vec{i}-22\vec{j}\\ &\textrm{E}.\quad 22\vec{i}-23\vec{j}\\\\ &\textrm{Jawab}:\qquad \textbf{D}\\ &\begin{aligned}\textrm{Ingat}&\textrm{lah akan tigaan Pythagoras}\\ &\begin{cases} (3,4,5) &\Rightarrow 3^{2}+4^{2}=5^{2} \\ (5,12,13) & \Rightarrow 5^{2}+12^{2}=13^{2} \\ (8,15,17) &\Rightarrow \cdots \\ (20,21,29)&\Rightarrow \cdots \\ \vdots & dll \end{cases}\\ \textrm{Sehin}&\textrm{gga}\\ \textrm{yang}&\: \: \textrm{paling mungkin adalah}\: : \: \\ &=\sqrt{20^{2}+21^{2}}=\sqrt{400+441}=\sqrt{841}\\ &=29 \end{aligned} \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Jika vektor}\: \: \vec{p}=\begin{pmatrix} 8\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} -3\\ 9 \end{pmatrix}\: \: \textrm{Hasil dari}\: \: \vec{p}+\vec{q}\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \begin{pmatrix} 6\\ 14 \end{pmatrix}\\ &\textrm{B}.\quad \begin{pmatrix} 6\\ 13 \end{pmatrix}\\ &\textrm{C}.\quad \begin{pmatrix} 6\\ 15 \end{pmatrix}\\ &\textrm{D}.\quad \begin{pmatrix} 5\\ 16 \end{pmatrix}\\ &\textrm{E}.\quad \begin{pmatrix} 5\\ 48 \end{pmatrix}\\\\ &\textrm{Jawab}:\qquad \textbf{D}\\ &\begin{aligned}\vec{p}+\vec{q}&=\begin{pmatrix} 8\\ 7 \end{pmatrix}+\begin{pmatrix} -3\\ 9 \end{pmatrix}\\ &=\begin{pmatrix} 8-3\\ 7+9 \end{pmatrix}\\ &=\begin{pmatrix} 5\\ 16 \end{pmatrix} \end{aligned} \end{array}.

\begin{array}{ll}\\ 5.&\textrm{Jika vektor}\: \: \vec{p}=\begin{pmatrix} ^{2}\log 32\\ -\, ^{3}\log \displaystyle \frac{1}{81} \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} -9\\ -21 \end{pmatrix}\: \: \textrm{Hasil dari}\: \: \vec{p}+\vec{q}\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \begin{pmatrix} 6\\ 17 \end{pmatrix}\\ &\textrm{B}.\quad \begin{pmatrix} -6\\ -17 \end{pmatrix}\\ &\textrm{C}.\quad \begin{pmatrix} 4\\ 17 \end{pmatrix}\\ &\textrm{D}.\quad \begin{pmatrix} -4\\ -17 \end{pmatrix}\\ &\textrm{E}.\quad \begin{pmatrix} 5\\ -16 \end{pmatrix}\\\\ &\textrm{Jawab}:\qquad \textbf{D}\\ &\begin{aligned}\vec{p}+\vec{q}&=\begin{pmatrix} ^{2}\log 32\\ -\, ^{3}\log \displaystyle \frac{1}{81} \end{pmatrix}+\begin{pmatrix} -9\\ -21 \end{pmatrix}\\ &=\begin{pmatrix} 5-9\\ 4-21 \end{pmatrix}\\ &=\begin{pmatrix} -4\\ -17 \end{pmatrix} \end{aligned} \end{array}.

\begin{array}{ll}\\ 6.&\textrm{Perhatikanlah gambar berikut}\\ & \end{array}.

\begin{array}{ll}\\ .\quad &\textrm{Panjang vektor}\: \: \vec{h}\: \: \textrm{tersebut di atas adalah}....\\ &\textrm{A}.\quad 5\\ &\textrm{B}.\quad 7\\ &\textrm{C}.\quad 10\\ &\textrm{D}.\quad 12\\ &\textrm{E}.\quad 15\\\\ &\textrm{Jawab}:\qquad \textbf{C}\\ &\begin{aligned}\textrm{Diketahui}&\quad \vec{h}=\overline{OH}=8\vec{i}+6\vec{j}\\ \vec{h}&=\sqrt{x_{H}^{2}+y_{H}^{2}}\\ &=\sqrt{8^{2}+6^{2}}\qquad \textbf{(ingat tripel Pythagoras)}\\ &=\sqrt{10^{2}}=10 \end{aligned} \end{array}.

\begin{array}{ll}\\ 7.&\textrm{Diketahui titik}\: \: P(n,2),\: Q(1,-2),\: n>0\: \: \textrm{dan panjang}\: \: \overline{PQ}=5,\: \: \textrm{maka nilai}\: \: n\: \: \textrm{adalah}....\\ &\textrm{A}.\quad 1\\ &\textrm{B}.\quad 2\\ &\textrm{C}.\quad 3\\ &\textrm{D}.\quad 4\\ &\textrm{E}.\quad 5\\\\ &\textrm{Jawab}:\qquad \textbf{D}\\ &\begin{aligned}\overline{PQ}&=5\\ \sqrt{(x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}}&=5\\ (x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}&=25\\ (1-n)^{2}+(-2-2)^{2}&=25\\ (1-n)^{2}+16&=25\\ (1-n)^{2}-3^{2}&=0\\ (1+3-n)(1-3-n)&=0\\ n=4\: \: \textrm{atau}\: \: n=-2& \end{aligned} \end{array}.

\begin{array}{ll}\\ 8.&\textrm{Diketahui vektor}\: \: \vec{u}=\begin{pmatrix} 3\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{v}=\begin{pmatrix} 2\\ -1 \end{pmatrix}.\: \textrm{Nilai}\: \: \left | \vec{u}+\vec{v} \right |\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \sqrt{28}\\ &\textrm{B}.\quad \sqrt{30}\\ &\textrm{C}.\quad \sqrt{34}\\ &\textrm{D}.\quad \sqrt{44}\\ &\textrm{E}.\quad \sqrt{50}\\\\ &\textrm{Jawab}:\qquad \textbf{C}\\ &\begin{aligned}\vec{u}+\vec{v}&=\begin{pmatrix} 3\\ 4 \end{pmatrix} + \begin{pmatrix} 2\\ -1 \end{pmatrix}\\ &=\begin{pmatrix} 3+2\\ 4+(-1) \end{pmatrix}\\ &=\begin{pmatrix} 5\\ 3 \end{pmatrix}\\ \left | \vec{u}+\vec{v} \right |&=\sqrt{5^{2}+3^{2}}\\ &=\sqrt{25+9}\\ &=\sqrt{34} \end{aligned} \end{array}.

\begin{array}{ll}\\ 9.&\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}....\\ &\textrm{A}.\quad \begin{pmatrix} 12\\ -16 \end{pmatrix}\\ &\textrm{B}.\quad \begin{pmatrix} 24\\ -16 \end{pmatrix}\\ &\textrm{C}.\quad \begin{pmatrix} 12\\ 16 \end{pmatrix}\\ &\textrm{D}.\quad \begin{pmatrix} 24\\ 16 \end{pmatrix}\\ &\textrm{E}.\quad \begin{pmatrix} -12\\ -16 \end{pmatrix}\\\\ &\textrm{Jawab}:\qquad \textbf{A}\\ &\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}\\ &=\begin{pmatrix} 12\\ -16 \end{pmatrix} \end{aligned} \end{array}.

Gambar berikut untuk soal No.10

\begin{array}{ll}\\ 10.&\textrm{Diketahui jajar genjang ABCD dengan titik E adalah perpotongan diagonal jajar genjang}.\\ &\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}....\\ &\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 \displaystyle -\frac{1}{2}\left ( \vec{a}+\vec{b} \right )\\ &\textrm{E}.\quad -\displaystyle \frac{1}{2}\left ( 2\vec{a}+\vec{b} \right )\\\\ &\textrm{Jawab}:\qquad \textbf{D}\\ &\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 )\\ &=-\displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right ) \end{aligned} \end{array}.

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Lanjutan Materi Vektor ( Kelas X Mat Pemintan)

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Vektor Dimensi Tiga (Materi Kelas X Matematika Peminatan)

 

Lihat Materi Sebelumnya, di sini

(Sebagai pengingat)

Vektor adalah besaran yang memiliki besar/panjang dan juga memiliki arah. Secara geometri atau grafis, vektor digambarkan dengan anak panah (ruas garis berarah) yang memiliki titik pangkal dan titik ujung.

Perhatikanlah ilustrasi berikut ini

Perhatikan salah satu ruas garis berarah pada gambar di atas, misal ruas garis berarah AB, dengan titik pangkal A dan titik ujungnya adalah B yang menyatakan sebagai vektor  \overrightarrow{AB} .

Berikut adalah cara penulisan notasi vektor

  • Dengan menggunakan dua huruf kapital yang ditasnya ada anak panah \overrightarrow{AB}  .
  • Dengan menggunakan dua huruf kapital yang di atasnya ada ruas garis, misal  \overline{AB}  .
  • Dengan menggunakan sebuah huruf kecil yang dicetak tebal, misal  \textbf{a},\: \textbf{b},\: \textbf{c},\: \textrm{dst}  .
  • Dengan menggunakan sebuah huruf kecil yang di atasnya ada, anak panah, Sebagai misal  \overrightarrow{a},\: \overrightarrow{b},\: \overrightarrow{c},\: \textrm{dst}  .
  • Dengan menggunakan huruf kecil yang di atas atau di bawahnya ada ruas garis. Sebagai contoh :  \overline{a},\: \overline{b},\: \overline{c},\: ...\: \textrm{atau}\: \underline{a},\: \underline{b},\: \underline{c},\: ...\: \textrm{dst}  .

A. Koordinat di Dimensi Tiga (Ruang Tiga, \textbf{R}^{3})

Perhatikanlah ilustrasi berikut

B. Vektro Basis

Sebagai contoh ilustrasi berikut:

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