$P$ is a point on $x+y+5=0$, whose perpendicular distance from $2 x+3 y+3=0$ is $\sqrt{13}$, then the coordinates of $P$ are:

Question

Aicharya · Accepted Answer

Let $P$ be $(x,-x-5)$
Perpendicular distance of $P$ from $2 x+3 y+3=0$ is $\sqrt{13}$ $\begin{aligned} & \Rightarrow\left|\frac{2 x-3 x-15+3}{\sqrt{13}}\right|=\sqrt{13} \Rightarrow|x+12|=13 \\ & \Rightarrow x=1,-25 \\ & x=1 \Rightarrow y=-6 \therefore \text { Required point is }(1,-6) \end{aligned}$

$P$ is a point on $x+y+5=0$, whose perpendicular distance from $2 x+3 y+3=0$ is $\sqrt{13}$, then the…

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