Question

The coordinates of the point(s) on $x+y+3=0$, whose distance from $x+2y+2=0$ is $\sqrt{5}$ units, is/are

• A. $(-9,6)$
• B. $(1,-4)$
• C. $(9,-6)$
• D. $(1,4)$

Answer 1

Answer: (A), (B)

$(1,-4)$

Let the points on the line $x+y+3=0$ is
$(a,-3-a)$

Now, length of the perpendicular from $(a,-3-a)$ to $x+2y+2=0$ is $\sqrt{5}$.
$\left|\frac{a-6-2a+2}{\sqrt{1^2+2^2}}\right|=\sqrt{5}\Rightarrow a=-9,1$

$\therefore (a,b)=(-9,6),(1,-4)$