Question

distance between A point (3,-4)and a line 3x-4y+10=0

Answer 1

Answer:

Step-by-step explanation:

Given,

Point 'D' = (3 , -4)

Line :- 3x - 4y + 10 = 0

To Find :-

Distance between the points

Solution :-

Comparing "3x - 4y + 10 = 0" with normal of form of line "Ax + By + C = 0" :-

A = 3 , B = - 4 , C = 10

D = (3 , -4) = ( x , y)

Formula Required :-

$d=\dfrac{|Ax+By+C|}{\sqrt{(A^2+B^2)}}$

Let's do :-

$d=\dfrac{|3(3)-4(-4)+10|}{\sqrt{(3)^2+(-4)^2}}$

$d=\dfrac{|9+16+10|}{\sqrt{9+16}}$

$d=\dfrac{|35|}{\sqrt{25}}$

$d=\dfrac{35}{5}$

d = 7

Distance between (3 , -4) and '3x - 4y + 10 = 0' is "7 units"