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Lanjutan Contoh Soal Transformasi Geometri

\begin{array}{ll}\\ \fbox{1}.&\textrm{Sebuah transformasi dengan}\: \: {x}'=2x+3y\: \: \textrm{dan}\: \: {y}'=3x+2y.\\ &\textrm{Peta titik}\: \: A(2,-1)\: \: \textrm{oleh transformasi tersebut adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (7,10)&&\textrm{d}.\quad (1,10)\\ \textrm{b}.\quad (10,7)&\textrm{c}.\quad (1,4)&\textrm{e}.\quad (4,1) \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{c}\\\\ \begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\begin{cases} {x}' & =2x+3y \\\\ {y}' & =3x+2y \end{cases}}\\\hline \begin{aligned}A(x,y)&\rightarrow {A}'({x}',{y}')\\ A(2,-1)&\rightarrow {A}'(1,4) \end{aligned}&\begin{aligned}{A}'({x}',{y}')&={A}'(2x+3y,\, 3x+2y)\\ &={A}'(2.2+3(-1),\, 3(2)+2(-1))\\ &={A}'(4-3,\, 6-2)\\ &={A}'(1,4) \end{aligned}\\\hline \end{array}.

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Lanjutan Contoh Soal Vektor 2

\begin{array}{ll}\\ \fbox{11}.&\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}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2&&\textrm{d}.\quad 8\\ \textrm{b}.\quad 4&\textrm{c}.\quad 6&\textrm{e}.\quad 10\end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

\begin{array}{ll}\\ \fbox{12}.&\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 =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{3}{5}&&\textrm{d}.\quad \displaystyle \frac{4}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{9}{16}&\textrm{c}.\quad \displaystyle \frac{3}{4}&\textrm{e}.\quad \displaystyle \frac{16}{9} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Jika}\: \: \vec{a}=\begin{pmatrix} 2\\ -5 \end{pmatrix},\: \vec{b}=\begin{pmatrix} 4\\ 3 \end{pmatrix},\: \textrm{maka proyeksi skalar ortogonal vektor}\: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{3}{5}&&\textrm{d}.\quad \displaystyle \frac{9}{5}\\\\ \textrm{b}.\quad \displaystyle \frac{7}{5}&\textrm{c}.\quad \displaystyle \frac{8}{5}&\textrm{e}.\quad \displaystyle \frac{10}{5} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{b}\\\\ \begin{aligned}\left | \vec{p} \right |&=\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{b} \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}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Vektor satuan untuk}\: \: \vec{a}=\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}\: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\sqrt{3}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}&&\textrm{d}.\quad \displaystyle \frac{1}{7}\sqrt{7}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}\\\\ \textrm{b}.\quad \displaystyle \frac{1}{3}\sqrt{3}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}&\textrm{c}.\quad \displaystyle \frac{1}{5}\sqrt{5}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}&\textrm{e}.\quad \displaystyle \frac{1}{21}\sqrt{21}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}\textrm{Vektor satuan}\: \: \vec{a}&\: \: \textrm{adalah}\: \: \vec{e}_{\vec{a}},\: \textrm{yaitu}:\\ \vec{e}_{\vec{a}}&=\displaystyle \frac{\vec{a}}{\left | \vec{a} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}}{\sqrt{2^{2}+(-1)^{2}+4^{2}}}\\ &=\displaystyle \frac{1}{\sqrt{21}}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}\\ &=\displaystyle \frac{1}{21}\sqrt{21}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix} \end{aligned}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Jika diketahui}\: \: \left | \vec{a} \right |=4\sqrt{3},\: \left | \vec{b} \right |=5,\: \textrm{ dan}\: \: \left ( \vec{a}+\vec{b} \right ).\left ( \vec{a}+\vec{b} \right )=13,\: \: \textrm{maka}\: \: \angle \left ( \vec{a},\: \vec{b} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 30^{\circ}&&\textrm{d}.\quad \displaystyle 135^{\circ}\\ \textrm{b}.\quad \displaystyle 60^{\circ}&\textrm{c}.\quad \displaystyle 120^{\circ}&\textrm{e}.\quad \displaystyle 150^{\circ} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}\left ( \vec{a}+\vec{b} \right ).\left ( \vec{a}+\vec{b} \right )&=13\\ \vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{b}.\vec{a}+\vec{b}.\vec{b}&=13\\ \left | \vec{a} \right |^{2}+2\vec{a}.\vec{b}+\left | \vec{b} \right |^{2}&=13,\qquad \textrm{ingat bahwa}\: \: \vec{a}.\vec{b}=\vec{b}.\vec{a}\\ \left ( 4\sqrt{3} \right )^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \angle \left ( \vec{a},\: \vec{b} \right )+5^{2}&=13\\ 48+2.(4\sqrt{3}).5.\cos \angle \left ( \vec{a},\: \vec{b} \right )+25&=13\\ 40\sqrt{3}\cos \angle \left ( \vec{a},\: \vec{b} \right )&=13-25-48\\ \cos \angle \left ( \vec{a},\: \vec{b} \right )&=\displaystyle \frac{-60}{40\sqrt{3}}\\ &=-\displaystyle \frac{1}{2}\sqrt{3}\\ &=-\cos 30\\ &=\cos \left ( 180^{\circ}-30^{\circ} \right )\\ \cos \angle \left ( \vec{a},\: \vec{b} \right )&=\cos 150^{\circ}\\ \angle \left ( \vec{a},\: \vec{b} \right )&=150^{\circ} \end{aligned}.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Jika diketahui titik}\: \: A(2,-1,4),\: B(4,1,3),\: \textrm{ dan}\: \: C(2,0,5),\: \: \textrm{maka}\: \: \sin \angle \left ( \overline{AB},\: \overline{AC} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{7}\sqrt{5}&&\textrm{d}.\quad \displaystyle \frac{1}{6}\sqrt{3}\\\\ \textrm{b}.\quad \displaystyle \frac{1}{6}\sqrt{34}&\textrm{c}.\quad \displaystyle \frac{2}{3}\sqrt{2}&\textrm{e}.\quad \displaystyle \frac{1}{6}\sqrt{2} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{b}\\\\ \begin{aligned}\cos \angle \left ( \overline{AB},\: \overline{AC} \right )&=\displaystyle \frac{\overline{AB}.\, \overline{AC}}{\left | \overline{AB} \right |.\left | \overline{AC} \right |} \\ &=\displaystyle \frac{(\vec{b}-\vec{a}).(\vec{c}-\vec{a})}{\sqrt{x^{2}_(\vec{b}-\vec{a})+y^{2}_(\vec{b}-\vec{a})+z^{2}_(\vec{b}-\vec{a})}.\sqrt{x^{2}_(\vec{c}-\vec{a})+y^{2}_(\vec{c}-\vec{a})+z^{2}_(\vec{c}-\vec{a})}} \\ &=\displaystyle \frac{\begin{pmatrix} 4-2\\ 1+1\\ 3-4 \end{pmatrix}.\begin{pmatrix} 2-2\\ 0+1\\ 5-4 \end{pmatrix}}{\sqrt{(4-2)^{2}+(1+1)^{2}+(3-4)^{2}}.\sqrt{(2-2)^{2}+(0+1)^{2}+(5-4)^{2}}}\\ &=\displaystyle \frac{2.0+2.1+-1.1}{\sqrt{4+4+1}.\sqrt{0+1+1}}\\ &=\displaystyle \frac{1}{3\sqrt{2}}\\ &=\displaystyle \frac{1}{6}\sqrt{2} \end{aligned}.

\begin{aligned}\textrm{Sehingga},\qquad \qquad &\\ \sin \angle \left ( \overline{AB},\: \overline{AC} \right )&=\sqrt{1-\cos ^{2}\angle \left ( \overline{AB},\: \overline{AC} \right )}\\ &=\sqrt{1-\left ( \displaystyle \frac{1}{6}\sqrt{2} \right )^{2}}\\ &=\sqrt{1-\displaystyle \frac{2}{36}}\\ &=\sqrt{\displaystyle \frac{34}{36}}\\ &=\displaystyle \frac{1}{6}\sqrt{34} \end{aligned}.

 

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Contoh Soal Trigonometri 2

\begin{array}{ll}\\ \fbox{11}.&\textrm{Dalam}\: \: \bigtriangleup \textrm{ABC}\: \: \textrm{berlaku}\: \: \sin \displaystyle \frac{1}{2}(\textrm{A+B})=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos \displaystyle \frac{1}{2}\textrm{C}&&\textrm{d}.\quad -\sin \displaystyle \frac{1}{2}\textrm{C}\\\\ \textrm{b}.\quad \sin \displaystyle \frac{1}{2}\textrm{C}&\textrm{c}.\quad -\cos \displaystyle \frac{1}{2}\textrm{C}&\textrm{e}.\quad \sin \displaystyle \textrm{C} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\sin \displaystyle \frac{1}{2}(\textrm{A+B})&=\sin \displaystyle \frac{1}{2}\left ( 180^{\circ}-\textrm{C} \right )\\ &=\sin \left ( 90^{\circ}-\displaystyle \frac{1}{2}\textrm{C} \right )\\ &=\cos \displaystyle \frac{1}{2}\textrm{C} \end{aligned}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Nilai}\: \: \cos 75^{\circ}\sin 15^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4}\left ( 2-\sqrt{3} \right )&&\textrm{d}.\quad \displaystyle \frac{1}{4}\left ( 1-\sqrt{3} \right )\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\left ( 2-\sqrt{3} \right )&\textrm{c}.\quad \displaystyle \frac{1}{2}\left ( 1-\sqrt{3} \right )&\textrm{e}.\quad \displaystyle \left ( 2-\sqrt{3} \right ) \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\cos 75^{\circ}\sin 15^{\circ}&=\displaystyle \frac{1}{2}\times \left ( 2\cos 75^{\circ}\sin 15^{\circ} \right )\\ &=\displaystyle \frac{1}{2}\left [\sin \left ( 75^{\circ}+15^{\circ} \right )-\sin \left ( 75^{\circ}-15^{\circ} \right ) \right ]\\ &=\displaystyle \frac{1}{2}\left [\sin 90^{\circ}-\sin 60^{\circ} \right ]\\ &=\displaystyle \frac{1}{2}\left [ 1-\frac{1}{2}\sqrt{3} \right ]\\ &=\displaystyle \frac{1}{4}\left [ 2-\sqrt{3} \right ] \end{aligned}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Nilai}\: \: 2\cos 50^{\circ}\cos 40^{\circ}-2\sin 95^{\circ}\sin 85^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 2\cos 10^{\circ}&&\textrm{d}.\quad 1\\\\ \textrm{b}.\quad \displaystyle 1+2\cos 10^{\circ}&\textrm{c}.\quad \displaystyle -1+2\cos 10^{\circ}&\textrm{e}.\quad -1 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}2\cos 50^{\circ}&\cos 40^{\circ}-2\sin 95^{\circ}\sin 85^{\circ}\\ &=\cos \left ( 50^{\circ}+40^{\circ} \right )+\cos \left ( 50^{\circ}-40^{\circ} \right )+\cos \left ( 95^{\circ}+85^{\circ} \right )-\cos \left ( 95^{\circ}-85^{\circ} \right )\\ &=2\cos 90^{\circ}+\cos 10^{\circ}+\cos 180^{\circ}-\cos 10^{\circ}\\ &=-1 \end{aligned}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Nilai}\: \: \displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -1&&\textrm{d}.\quad \displaystyle \frac{1}{3}\sqrt{6}\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\sqrt{2}&\textrm{c}.\quad 1&\textrm{e}.\quad \displaystyle \frac{1}{2}\sqrt{6} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}&=\displaystyle \frac{-2\sin \displaystyle \frac{1}{2}\left ( 75^{\circ}+15^{\circ} \right )\sin \frac{1}{2}\left ( 75^{\circ}-15^{\circ} \right )}{2\cos \displaystyle \frac{1}{2}\left ( 75^{\circ}+15^{\circ} \right )\sin \frac{1}{2}\left ( 75^{\circ}-15^{\circ} \right )}\\ &=-\displaystyle \frac{2\sin 45^{\circ}\times \sin 30^{\circ}}{2\cos 45^{\circ}\times \sin 30^{\circ}}\\ &=-\tan 45^{\circ}\\ &=-1 \end{aligned}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Bentuk sederhana dari}\: \: 2\cos \left ( x+\displaystyle \frac{\pi }{4} \right ). \cos \left ( \displaystyle \frac{3}{4}\pi -x \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos (2x+\pi )&&\textrm{d}.\quad -\cos \left ( 2x-\displaystyle \frac{\pi }{2} \right )\\\\ \textrm{b}.\quad \cos \left ( 2x+\displaystyle \frac{\pi }{2} \right )&\textrm{c}.\quad \cos 2x&\textrm{e}.\quad \sin 2x-1 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}2\cos \left ( x+\displaystyle \frac{\pi }{4} \right ). \cos \left ( \displaystyle \frac{3}{4}\pi -x \right )&=\cos \left ( x+\displaystyle \frac{\pi }{4}+\displaystyle \frac{3}{4}\pi -x \right )+\cos \left (x+\displaystyle \frac{\pi }{4} -\left ( \displaystyle \frac{3}{4}\pi -x \right ) \right )\\ &=\cos \pi +\cos \left ( 2x-\frac{2}{4}\pi \right )\\ &=-1+\cos \left ( \displaystyle \frac{1}{2}\pi -2x \right ),\quad\quad \textrm{ingat bahwa}\: \cos (-\alpha )=\cos \alpha \\ &=-1+\sin 2x\\ &=\sin 2x-1 \end{aligned}.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Nilai dari}\: \: \cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1&&\textrm{d}.\quad -\displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -1&\textrm{c}.\quad \displaystyle \frac{1}{2}&\textrm{e}.\quad 0\end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}\cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}&=2\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}+40^{\circ} \right )\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}-40^{\circ} \right )-\cos 20^{\circ}\\ &=2\cos 60^{\circ}\cos 20^{\circ}-\cos 20^{\circ}\\ &=2.\displaystyle \frac{1}{2}.\cos 20^{\circ}-\cos 20^{\circ}\\ &=\cos 20^{\circ}-\cos 20^{\circ}\\ &=0 \end{aligned}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Bentuk sederhana dari}\: \: 2\sin \left ( \displaystyle \frac{3}{4}\pi +x \right ).\cos \left ( \displaystyle \frac{\pi }{4}+x \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1-\sin 2x&&\textrm{d}.\quad \cos 2x\\\\ \textrm{b}.\quad 1+\sin 2x&\textrm{c}.\quad 1-\cos 2x&\textrm{e}.\quad -\cos 2x\end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}2\sin \left ( \displaystyle \frac{3}{4}\pi +x \right ).\cos \left ( \displaystyle \frac{\pi }{4}+x \right )&=2\sin \left ( 135^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\sin \left ( 90^{\circ}+45^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\cos \left ( 45^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\cos ^{2}\left ( 45^{\circ}+x \right )\\ &=2\left ( \cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x \right )^{2}\\ &=2\left ( \frac{1}{2}\sqrt{2}.\cos x-\displaystyle \frac{1}{2}\sqrt{2}.\sin x \right )^{2}\\ &=\displaystyle \left ( \cos x-\sin x \right )^{2}\\ &=\cos ^{2}x+\sin ^{2}x-2\sin x\cos x\\ &=1-\sin 2x \end{aligned}.

Sumber Referensi

  1. Kartini, Suprapto, Subandi, dan Untung Setiyadi. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
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Lanjutan Contoh Soal Vektor (Kelas XII IPA KTSP)

\begin{array}{ll}\\ \fbox{1}.&\textrm{Panjang vektor}\: \: \vec{a}=\displaystyle \begin{pmatrix} -7\\ 24 \end{pmatrix}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 8&&\textrm{d}.\quad 25\\ \textrm{b}.\quad 12&\textrm{c}.\quad 17&\textrm{e}.\quad 35 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{d}\\\\ \begin{aligned}\textrm{Panjang Vektor}\: \: \vec{a}&=\displaystyle \begin{pmatrix} -7\\ 24 \end{pmatrix}\: \: \textrm{adalah}:\\ \left | \vec{a} \right |&=\sqrt{(-7)^{2}+(24)^{2}}\\ &=\sqrt{49+576}\\ &=\sqrt{625}\\ &=25 \end{aligned}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Diketahui titik}\: \: P(n,2),\: Q(1,-2),\: n>0\: \: \textrm{dan panjang}\: \: \overline{PQ}=5,\: \: \textrm{maka nilai}\: \: n\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1&&\textrm{d}.\quad 4\\ \textrm{b}.\quad 2&\textrm{c}.\quad 3&\textrm{e}.\quad 5 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui vektor}\: \: \vec{a}=\begin{pmatrix} 3\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 2\\ -1 \end{pmatrix}.\: \textrm{Nilai}\: \: \left | \vec{a}+\vec{b} \right |\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{28}&&\textrm{d}.\quad \sqrt{44}\\ \textrm{b}.\quad \sqrt{30}&\textrm{c}.\quad \sqrt{34}&\textrm{e}.\quad \sqrt{50} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{c}\\\\ \begin{aligned}\vec{a}+\vec{b}&=\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{a}+\vec{b} \right |&=\sqrt{5^{2}+3^{2}}\\ &=\sqrt{25+9}\\ &=\sqrt{34} \end{aligned}.

\begin{array}{ll}\\ \fbox{4}.&\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 \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} \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

Gambar berikut untuk soal No.5

374

\begin{array}{ll}\\ \fbox{5}.&\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}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right )&&\textrm{d}.\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{e}.\quad -\displaystyle \frac{1}{2}\left ( 2\vec{a}+\vec{b} \right ) \end{array} \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

Gambar berikut untuk soal No.6

373

\begin{array}{ll}\\ \fbox{6}.&\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{d}.\quad \displaystyle \frac{1}{3}\left ( \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{e}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-\vec{q} \right ) \end{array} \end{array}\\\\\\ \textrm{Jawab}:\: \: \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} .

\begin{array}{ll}\\ \fbox{7}.&\textrm{Vektor}\: \: \vec{v}=\begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{searah dengan vektor}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \begin{pmatrix} 2\\ -5 \end{pmatrix}&&\textrm{d}.\quad \displaystyle \begin{pmatrix} -4\\ 5 \end{pmatrix}\\ \textrm{b}.\quad \displaystyle \begin{pmatrix} 2\\ 5 \end{pmatrix}&\textrm{c}.\quad \displaystyle \begin{pmatrix} -6\\ 15 \end{pmatrix}&\textrm{e}.\quad \displaystyle \begin{pmatrix} -3\\ 10 \end{pmatrix} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{c}\\\\ \begin{aligned}\textrm{Vektor}&\: \: \vec{v}\: \: \: \textrm{searah dengan vektor}\: \: k.\vec{v}\\\\ &\begin{array}{|c|c|c|c|c|}\hline \multicolumn{5}{|c|}{k.\vec{v}=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}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Vektor satuan}\: \: \vec{v}=\begin{pmatrix} -5\\ -12 \end{pmatrix}\: \: \textrm{adalah}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -\frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix}&&\textrm{d}.\quad \displaystyle -\frac{1}{17}\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{e}.\quad \displaystyle \frac{1}{2} \begin{pmatrix} 5\\ 12 \end{pmatrix} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\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}.

\begin{array}{ll}\\ \fbox{9}.&\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}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} 4\\ 4 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} 4\\ 2 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 4\\ 5 \end{pmatrix}&\textrm{c}.\quad \begin{pmatrix} -4\\ 4 \end{pmatrix}&\textrm{e}.\quad \begin{pmatrix} -4\\ 6 \end{pmatrix} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Jika}\: \: \vec{a}=\begin{pmatrix} -2\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 8\\ 4 \end{pmatrix},\: \: \textrm{maka sudut yang dibentuk vektor}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 0^{\circ}&&\textrm{d}.\quad 60^{\circ}\\ \textrm{b}.\quad 30^{\circ}&\textrm{c}.\quad 45^{\circ}&\textrm{e}.\quad 90^{\circ} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}\vec{a}.\vec{b}&=\displaystyle \begin{vmatrix} \vec{a} \end{vmatrix}.\begin{vmatrix} \vec{b} \end{vmatrix}.\cos \angle \left (\vec{a},\, \vec{b} \right )\\ \cos \angle \left (\vec{a},\, \vec{b} \right )&=\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |.\left | \vec{b} \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{a},\, \vec{b} \right )&=\cos 90^{\circ}\\ \angle \left (\vec{a},\, \vec{b} \right )&=90^{\circ} \end{aligned}.

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Lanjutan Contoh Soal Persamaan Lingkaran dan Garis Singgung 2

\begin{array}{ll}\\ \fbox{11}.&\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 \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}\\ (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}.

\begin{array}{ll}\\ \fbox{12}.&\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\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \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} (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}.

\begin{array}{ll}\\ \fbox{13}.&\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\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

\begin{array}{ll}\\ \fbox{14}.&\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\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

Perhatikanlah gambar berikut sebagai ilustrasinya

372

\begin{array}{ll}\\ \fbox{15}.&\textrm{Salah satu garis singgung lingkaran}\: \: x^{2}+y^{2}=10\: \: \textrm{yang ditarik dari}\\ &\textrm{titik}\: \: (4,2)\: \: \textrm{adalah}....\\ &\textrm{a}.\quad 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\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\.

\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 \multicolumn{2}{|c|}{\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 x+3y=10 \end{cases} \end{aligned}}\\\hline \end{array}.

Sumber Referensi

  1. Kartini, Suprapto, Subandi, dan Untung Setiyadi. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
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Contoh Soal Persamaan Lingkaran dan Garis Singgung (Kelas XI)

\begin{array}{ll}\\ \fbox{1}.&\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\\ \end{array}\\\\\\ \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}.

\begin{array}{ll}\\ \fbox{2}.&\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\\ \end{array}\\\\\\ \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{Alternatif 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}\\\\ \textrm{Silahkan dilengkapi sendiri}.

\begin{array}{ll}\\ \fbox{3}.&\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\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

\begin{array}{ll}\\ \fbox{4}.&\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\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

\begin{array}{ll}\\ \fbox{5}.&\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)\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{c}\\\\ \begin{aligned}\textrm{Lingkaran}\: \: x^{2}+y^{2}+px+8y+9&=0\: \: \textrm{menyinggung sumbu X} \: (\textrm{sejajar}\: \: y=0)\\ \textrm{maka,}\quad\quad\qquad\qquad\qquad\quad\quad\quad\: \: \: &\\ y=0\: \Rightarrow \: x^{2}+0^{2}+px+8.0+9&=0\\ x^{2}+px+9&=0\begin{cases} a & =1 \\ b & =p \\ c & =9 \end{cases}\: ,\: \textrm{syarat menyinggung}\: \: D=b^{2}-ac=0\\ b^{2}-4ac&=0\\ p^{2}-4(1)(9)&=0\\ p^{2}-36&=(p+6)(p-6)=0\\ x&=-6\: \: \textrm{atau}\: \: x=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}.

Sebagai ilustrasi perhatikanlah gambar berikut ini

365

\begin{array}{ll}\\ \fbox{6}.&\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)\\\end{array}\\\\\\ \textrm{Jawab}:\\\\ \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}.

Sebagai ilustrasi perhatikanlah gambar berikut:

366

\begin{array}{ll}\\ \fbox{7}.&\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\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

Berikut ilustrasi gambarnya;

367

\begin{array}{ll}\\ \fbox{8}.&\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}\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

Berikut ilustrasinya,

368

\begin{array}{ll}\\ \fbox{9}.&\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} \end{array}\\\\\\ \textrm{Jawab}:\: \: \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}.

\begin{array}{ll}\\ \fbox{10}.&\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 \end{array}\\\\\\ \textrm{Jawab}:\: \: \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 berada 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}.

Berikut ilustrasi gambarnya

369

Sumber Referensi

  1. Kartini, Suprapto, Subandi, dan Untung Setiyadi. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
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