#### Explain solution RD Sharma class 12 chapter Determinants exercise 5.3 question 12 subquestion (ii) maths

$\dpi{100} \boldsymbol{Answer\! :} x=3y$

$\dpi{100} \boldsymbol{Hints\! :} Take\; the\; third\; point\; as\; (x,y)\; and\;\; use\; the\; f\! ormula\; o\! f\; area\; o\! f\; triangle\; and\; equate\; it\; with\; zero.$

Given: (3,1) and (9,3)

Explanation: Vertices are (3,1) and (9,3)
$\dpi{100} Let\; the\; third\; point\; be\; (x,y)$

$\dpi{100} Determinant= \Delta =0$

$\dpi{100} \Delta =\frac{1}{2}\begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}=0$

$\dpi{100} \Rightarrow \frac{1}{2}\begin{vmatrix} x &y &1 \\ 3 &1 &1 \\ 9 &3 &1 \end{vmatrix}=0$

$\dpi{100} \Rightarrow x\left|\begin{array}{ll} 1 & 1 \\ 3 & 1 \end{array}\right|-y\left|\begin{array}{cc} 3 & 1 \\ 9 & 1 \end{array}\right|+1\left|\begin{array}{cc} 3 & 1 \\ 9 & 3 \end{array}\right|=0$

$\dpi{100} \Rightarrow x(1-3)-y(3-9)+1(9-9)=0$

$\dpi{100} \Rightarrow -2x+6y=0$

$\dpi{100} \Rightarrow -x+3y=0$

$\dpi{100} \Rightarrow x=3y$