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Q.
The distance of the point $(3,-5)$ from the line $3 x-4 y-26=0$ is
Straight Lines
Solution:
Given line is $3 x-4 y-26=0 ....$(i)
On comparing Eq. (i) with general equation of line
$A x+B y+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
$d =\frac{\left|A x_1+B y_1+C\right|}{\sqrt{A^2+B^2}}$
$ =\frac{|3 \cdot 3+(-4)(-5)-26|}{\sqrt{3^2+(-4)^2}}=\frac{3}{5}$