Question

The distance of the point $(3,-5)$ from the line $3x-4y-26=0$ is

• A. $\frac{3}{7}$
• B. $\frac{2}{5}$
• C. $\frac{7}{5}$
• D. $\frac{3}{5}$

Answer 1

Answer: (D)

Given line is $3x-4y-26=\mathrm{0....}$(i)
On comparing Eq. (i) with general equation of line
$Ax+By+C=0$, we get
$A=3,B=-4$
and $C=-26$
Given point is $\left(x_1,y_1\right)=(3,-5)$. The distance of the given point from given line is
$d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$
$=\frac{|3\cdot 3+(-4)(-5)-26|}{\sqrt{3^2+(-4{)}^2}}=\frac{3}{5}$