The reflection of the point \((4,-13)\) in the line \(5 x+y+6=0\), is
The reflection of the point \((4,-13)\) in the line \(5 x+y+6=0\), is
\((-1,-14)\)
\((3,4)\)
\((1,2)\)
\((-4,13)\)
Solution
Let $Q(a, b)$ be the reflection of $P(4,-13)$ in the line $5x+y+6=0$.
Then the mid-point $R\left(\frac{a+4}{2}, \frac{b-13}{2}\right)$ lies on
$\begin{aligned}
& 5x+y+6=0 \\
& \therefore 5\left(\frac{a+4}{2}\right)+\frac{b-13}{2}+6=0 \\
& \Rightarrow 5a+b+19=0 \quad ...(i)
\end{aligned}$
Also $PQ$ is perpendicular to $5x+y+6=0$. Therefore $\frac{b+13}{a-4} \times\left(-\frac{5}{1}\right)=-1$
$\Rightarrow a-5b-69=0 \quad ...(ii)$
Solving (i) and (ii), we get $a=-1, b=-14$