Find the point on the line −4x+2y+4=0 which is closest to the point (1,−5).

Answer 1

If (a,b) is a point not on Ax + By + C = 0
then the shortest distance D from (a,b) to the line is

D = |aA + bB + C|/√(A^2 + B^2)

First change your equation into general form:
4x - 2y - 4 = 0

D = |1(4) + (-5)(-2) - 4|/√(16+4)
= 10/√20
= 10/2√5
= 5/√5
= √5