Question

Number of points having distance $\sqrt{5}$ from the straight line $x-2y+1=0$ and a distance $\sqrt{13}$ from the line $2x+3y-1=0$ is

• A. $1$
• B. $2$
• C. $4$
• D. $5$

Answer 1

Answer: (C)

$4$

Let the point be $(a,b)$

$L_1:x-2y+1=0$

$L_2:2x+3y-1=0$

We know that the distance $d$ of a point $P(p,q)$, from a line $L:mx+ny+r=0$ can be calculated by:

$d=\left|\frac{pm+nq+r}{m^2+n^2}\right|$

Using this, we get

For $L_1$: $d=\sqrt{5}$
$\Rightarrow \left|\frac{a-2b+1}{\sqrt{5}}\right|=\sqrt{5}\Rightarrow a-2b+1=\pm 5$ ....(1)

For $L_2:d=\sqrt{13}$

$\Rightarrow \left|\frac{2a+3b-1}{\sqrt{13}}\right|=\sqrt{13}\Rightarrow 2a+3b-1=\pm 13$ .... (2)

Any one of equations $\left(1\right)$ combined with any one of equation $\left(2\right)$ will yield a point $\Rightarrow 4$ points