Question

find the distance of the line 4x+3y+5=0 from the origin​

Answer 1

Given :

Equation of line : 4x+3y+5=0

Point = (0,0)  

To find :

The distance of the line 4x+3y+5=0 from the origin​(0,0).

Solution :

The distance between the point (x,y) and line Ax+By+C=0 is given by:

$d = |A.x + B.y + C| / \sqrt{A^{2} + B^{2} }\\d = |4*0 +3*0 + 5| / \sqrt{4^{2} + 3^{2} } \\d = |5| / \sqrt{16 + 9}\\d = 5 / \sqrt{25} \\d = 5 / 5\\d = 1$

Therefore the distance of the line 4x+3y+5=0 from the origin (0,0) is 1unit.