Question

The coordinates of a point on the line $x+y+1=0$ which is at a distance $\frac{1}{5}$ unit from the line $3x+4y+2=0$ are

• A. $(2,-3)$
• B. $(-3,2)$
• C. $(0,-1)$
• D. $(-1,0)$

Answer 1

Answer: (B), (D)

The correct options are
B $(-3,2)$
D $(-1,0)$

Any point on the line $x+y+1=0$,
$=(-t-1,t)$
Now, distance of this coordinate from the line $3x+4y+2=0$ is given as $\frac{1}{5}$

$|\frac{3(-1-t)+4t+2}{\sqrt{3^2+4^2}}|=\frac{1}{5}$
$\Rightarrow |\frac{t-1}{\sqrt{25}}|=\frac{1}{5}$
$\Rightarrow \frac{|t-1|}{5}=\frac{1}{5}$
$\Rightarrow |t-1|=1$
$\Rightarrow t-1=\pm 1\Rightarrow t=2\text{or}0$
Thus, putting the values of $t$ in the parametric form of point on the line, we get coordinates as
$(-3,2)\text{and}(-1,0)$