Question

find the distance of point 3,-5 from the line 3x-4y-26=0

Answer 1

i hope it helps you. Plz mark my answer to brainliest

Answer 2

Answer:

The distance of point 3, -5 from the line is 3/5.

Step-by-step explanation:

Consider the given line 3x-4y-26=0

Compare the given line with ax+by+c=0

We get  

a=3 b=-4 c=-26

Given point= (3,-5)

Where x=3 and y=-5

We know that the distance from the point p to the line ax+by+c=0 is given by

$\bold{d=\left|\left[\frac{a x+b y+c}{\sqrt{a^{2}+b^{2}}}\right]\right|}$

By substituting the values for a, b, c, x, y we get

$d=\frac{[3(3)+(-4)(-5)-26]}{\sqrt{3^{2}+(-4)^{2}}}$

Simplify the above expression

$d=\frac{9+20-26}{\sqrt{25}}$

d=3/5

Distance from the point p to the line 3x-4y-26=0 is 3/5