Question

The image of the point $(-8,12)$ with respect to the line mirror $4x+7y+13=0$ is

• A. $(-16,2)$
• B. $(16,-2)$
• C. $(-16,-2)$
• D. $(-2,-16)$

Answer 1

The correct option is C $(-16,-2)$

Let the image of the point $P(-8,12)$ in the line mirror $AB$ be $Q(\alpha ,\beta )$ and mid point of $PQ$ be $M.$
Then, coordinates of $M$ are $(\frac{\alpha -8}{2},\frac{\beta +12}{2})$.


Since, $M$ lies on $4x+7y+13=0$,
$\therefore 2\alpha -16+\frac{7\beta +84}{2}+13=0\Rightarrow 4\alpha +7\beta +78=0\cdots \left(1\right)$
Since, $PQ$ is perpendicular to $AB$,
therefore, Slope of $AB\times$ Slope of $PQ$ $=-1$
$\Rightarrow (-\frac{4}{7})\times \left(\frac{\beta -12}{\alpha +8}\right)=-1$
$\Rightarrow 7\alpha -4\beta +104=0\cdots \left(2\right)$

On solving equation $\left(1\right)$ and $\left(2\right)$, we get
$\alpha =-16,\beta =-2$
Hence, the image of $(-8,12)$ in the line mirror $4x+7y+13=0$ is $(-16,-2)$

Alternate Solution :
Using the image formula
$\frac{x+8}{4}=\frac{y-12}{7}=-\frac{2(4\cdot (-8)+7\cdot 12+13)}{4^2+7^2}\Rightarrow x=-16;y=-2$
So image will be $(-16,-2)$