Question

The point on the line $x+y+3=0$ whose distance from $x+2y+2=0$ is $\sqrt{5}$

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

Answer 1

Answer: (D)

$(-9,6)$

Given

$x+y+3=0$

Let $(\alpha ,\beta )$ be a point

then $\alpha +\beta +3=0$

$\beta =-(\alpha +3)$

then $(\alpha ,\beta )=(\alpha ,-(\alpha +3\left)\right)$

Now let us find the

distance between $(\alpha ,-(\alpha +3\left)\right)$ and $x+2y+2=0$

$\left|\frac{\alpha +2[-(\alpha +3\left)\right]+2}{\sqrt{1^2+2^2}}\right|=\sqrt{5}$

$|\alpha -2\alpha -6+2|=\sqrt{5}.\sqrt{5}$

$|-\alpha -4|=5$

$\alpha +4=5$ ; $\alpha +4=-5$

$\alpha =1$ $\alpha =-9$

Then point will be

$(\alpha ,-(\alpha +3\left)\right)$

When $\alpha =1\Rightarrow (1,-(1+3\left)\right)$

$=(1,-4)$

$\alpha =-9\Rightarrow (-9,-(-9+3\left)\right)$

$=(-9,6)$