The co-ordinates of point on the line $x+y+3=0$, whose distance from the line $x+2 y+2=0$ is $\sqrt{5}$…
The co-ordinates of point on the line $x+y+3=0$, whose distance from the line $x+2 y+2=0$ is $\sqrt{5}$ units, are
(-1,-4)
(1,-4)
(-1,4)
(1,4)
Solution
Any point on the line $x+y+3=0$ can be taken as $(k,-3-k)$ Now, its distance from $x+2 y+2=0$ is
$\begin{aligned}
& \frac{|k+2 \times(-3-k)+2|}{\sqrt{1^2+2^2}}=\sqrt{5} \\
& \Rightarrow 1-k-41=5 \\
& \Rightarrow-k-4= \pm 5 \\
& \Rightarrow k=1,-9
\end{aligned}$
for $k=1$ the point is $(1,-4)$