Solusi Soal Nomor 3 Uji Sertifikasi Pendidikan Menengah

$latex 3.\ \ \ (a)\ \ x^2+y^2-4x+6y=87\\\begin{matrix} & & & \ \end{matrix}x^2+y^2-2(2)x-2(-3)y=87\\\begin{matrix} & & & \ \end{matrix}Ini\ menunjukkan\ persamaan\ lingkaran\ yang\ berpusat\ di\ titik\ (2,\ -3).\\\\\begin{matrix} & & & \ \end{matrix}(i)\ \ x+y+1=0\\\begin{matrix} & & & \ \end{matrix}\qquad Untuk\ titik\ (2,\ -3)\ diperoleh:\\\begin{matrix} & & & \ \end{matrix}\qquad \begin{array}{rcl}2+(-3)+1&=&0\\0&=&0\end{array}\\\begin{matrix} & & & \ \end{matrix}\qquad Karena\ koordinat\ titik\ pusat\ lingkaran\ memenuhi\ persamaan\ garis\\\begin{matrix} & & & \ \end{matrix}\qquad maka\ garis\ tersebut\ melewati\ titik\ pusat\ lingkaran.\\\\\begin{matrix} & & & \ \end{matrix}(ii)\ \ x^2+y^2-4x+6y=87\\\begin{matrix} & & & \ \end{matrix}\qquad {\bf x_{1}x+y_{1}y-2(x_{1}+x)+3(y_{1}+y)=87}\quad (x_{1}=-6\ dan\ y_{1}=3)\\\begin{matrix} & & & \ \end{matrix}\qquad -6x+3y-2(-6+x)+3(3+y)=87\\\begin{matrix} & & & \ \end{matrix}\qquad -6x+3y+12-2x+9+3y=87\\\begin{matrix} & & & \ \end{matrix}\qquad -8x+6y+21=87\\\begin{matrix} & & & \ \end{matrix}\qquad -8x+6y+21-87=0\\\begin{matrix} & & & \ \end{matrix}\qquad -8x+6y-66=0\\\begin{matrix} & & & \ \end{matrix}\qquad -4x+3y-33=0\\\begin{matrix} & & & \ \ \end{matrix}\qquad 4x-3y+33=0$ $latex \begin{matrix} & \ \ \end{matrix}(b)\ \ Karena\ \overrightarrow{AP}=\frac{1}{2}\ \overrightarrow{OA}\ maka\ \overrightarrow{AP}\ searah\ dengan\ \overrightarrow{OA}\ dan\ titik\ P\ terletak\\\begin{matrix} & & & \ \end{matrix}pada\ perpanjangan\ \overline{OA},\ sehingga:\\\\\begin{matrix} & & & \ \end{matrix}{\bf\overrightarrow{OP}=\frac{3}{2}\ \overrightarrow{OA}=\frac{3}{2}\ a}\\\\\begin{matrix} & & & \ \end{matrix}(i)\\\begin{matrix} & & & \ \end{matrix}\qquad \begin{array}{rcl}{\bf \overrightarrow{BP}}&=&{\bf \overrightarrow{BO}}+{\bf \overrightarrow{OP}}\\&=&{\bf -b+\frac{3}{2}\ a}\\&=&{\bf \frac{3}{2}\ a-b}\end{array}\\\\\begin{matrix} & & & \ \end{matrix}(ii)\\\begin{matrix} & & & \ \end{matrix}\qquad \begin{array}{rcl}{\bf \overrightarrow{BP}}&=&{\bf \frac{3}{2}\ a-b}\\\\&=&\frac{3}{2}\ \left ( \begin{matrix} 2\\ 1 \end{matrix} \right )-\left ( \begin{matrix} 3\\ 2 \end{matrix} \right )\\\\&=&\left ( \begin{matrix} 3\\ \frac{3}{2} \end{matrix} \right )-\left ( \begin{matrix} 3\\ 2 \end{matrix} \right )\\\\&=&\left ( \begin{matrix} 0\\ -\frac{1}{2} \end{matrix} \right )\end{array}\\\\\\\begin{matrix} & & & \ \end{matrix}\qquad \begin{array}{rcl}\big|{{\bf \overrightarrow{BP}}}\big|&=&\sqrt{0^2+\big(-\frac{1}{2}\big)^2}\\\\&=&\sqrt{0+\frac{1}{4}}\\\\&=&\sqrt{\frac{1}{4}}\\\\&=&\frac{1}{2}\end{array}$