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The reflection of the point $(4,-13)$ in the line $5 x+y+6=0$ is
Option: 1
$(-1,-14)$

Option: 2
$(3,4)$

Option: 3
$(1,2)$

Option: 4
$\quad(-4,13)$

Answers (1)

best_answer

Let Q(a, b) be the reflection of $P(4,-13)$ in the line $5 x+y+6=0$ . Then the point

$R\left(\frac{a+4}{2}, \frac{b-13}{2}\right) \text { lies on } 5 x+y+6=0 \quad$

$\therefore \quad 5\left(\frac{a+4}{2}\right)+\left(\frac{b-13}{2}\right)+6=0 \quad \Rightarrow 5 a+b+19=0$ $....(i)$

Also $P Q$ is perpendicular to $5 x+y+6=0$ .

Therefore $\left(\frac{b+13}{a-4}\right) \times\left(\frac{-5}{1}\right) \Rightarrow a-5 b-69=0$ $....(ii)$

Solving (i) and (ii), we get

$a=-1, b=-14$

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