$Q \equiv\left( x _{2}, mx _{2}\right)$
$A _{1}=\frac{1}{2}\left|\begin{array}{ccc}3 & 4 & 1 \\ 2 & 0 & 1 \\ -1 & 1 & 1\end{array}\right|=\frac{13}{2}$
$A _{2}=\frac{1}{2}\left|\begin{array}{ccc} x _{1} & mx _{1} & 1 \\ x _{2} & mx _{2} & 1 \\ 2 & 0 & 1\end{array}\right|$
$A _{2}=\frac{1}{2}\left|2\left( mx _{1}- mx _{2}\right)\right|= m \left| x _{1}- x _{2}\right|$
$A _{1}=3 A _{2} \Rightarrow \frac{13}{2}=3 m \left| x _{1}- x _{2}\right|$
$AC : x +3 y =2$
$BC : y =4 x -8$
$P : x +3 y =2$ and $y = mx \Rightarrow x _{1}=\frac{2}{1+3 m }$
$Q : y =4 x -8$ and $y = mx \Rightarrow x _{2}=\frac{8}{4- m }$
$\left| x _{1}- x _{2}\right|=\left|\frac{2}{1+3 m }-\frac{8}{4- m }\right|$
$=\left|\frac{-26 m }{(1+3 m )(4- m )}\right|=\frac{26 m }{(3 m +1)| m -4|}$
$=\frac{26 m }{(3 m +1)(4- m )}$
$\left| x _{1}- x _{2}\right|=\frac{13}{6 m }$
$\Rightarrow \frac{26 m }{(3 m +1)(4- m )}=\frac{13}{6 m }$
$\Rightarrow \quad 12 m ^{2}=-(3 m +1)( m -4)$
$\Rightarrow \quad 12 m ^{2}=-\left(3 m ^{2}-11 m -4\right)$
$\Rightarrow \quad 15 m ^{2}-11 m -4=0$
$\Rightarrow \quad 15 m ^{2}-15 m +4 m -4=0$
$\Rightarrow \quad(15 m +4)( m -1)=0$
$\Rightarrow m =1$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(a)$ reflection about the line $y=x$.
$(b)$ translation through $2$ units along the positive direction of $x$-axis.
$(c)$ rotation through angle $\frac{\pi}{4}$ about the origin in the anti-clockwise direction.
If the co-ordinates of the final position of the point $P$ are $\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$, then the value of $2 a+b$ is equal to: