The distance between the point (2,-3) and the line 4x+3y-C = 0 is 5 units. Find C.

The distance between the point (x1 , y1) and the line ax +
by + c = 0 is given by |a*x1 + b*y1 + c|/sqrt (a^2 +
b^2)

Substituting the values we have
here

D = 5 = |4*2 + 3*(-3) - C|/sqrt (16 +
9)

=> 5 = |8 - 9 - C| /
5

=> 25 = |-1 - C|

-25
= 1 + C

=> C = -26

Also
25 = -1 - C

=> C =
24

The values of C are C = -26 and C=
24