Question

The distance between two parallel lines $3x+4y+10=0$ and $3x+4y-10=0$ is

• A. $0$
• B. $-4\sqrt{5}$
• C. $2\sqrt{5}$
• D. $4$

Answer 1

The correct option is D $4$

The equation of one line is $3x+4y+10=0$

Putting $x=0$, we get

$4y=-10$

$\Rightarrow y=-\frac{5}{2}$

Therefore, the point of intersection with $Y$-axis is

$A=(x_1,y_1)=(0,-\frac{5}{2})$.

The equation of the second line is

$3x+4y-10=0$ equivalent to $ax+by+c=0$

Therefore, the perpendicular distance of $A$ from this line is

$|p|=\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}=\frac{3\times 0+4\times (-\frac{5}{2})-10}{\sqrt{3^2+4^2}}=-4$

$\Rightarrow p=4$ ..... [Since $p$ is a length, we take the positive value ]