Find the perpendicular distance from the point (5, 6) to the line -2x + 3y + 4 = 0

The line prpendicular line to line ax+by+c = 0 is given
by:

 bx-ay+d = 0, where d is a varying constant and to be determined
by a condition.

Therefore the perpendicular line  which passes
through a fixed point (x1,y1) is given by:

a(x-x) -a(y-y1) =
0....(1)

The equation of the given line is 2x+3y+4 = 0. So any  line
perpendicular to this line is of the form 3x-2y +k = 0. Since this passes through the point
(5,6), it should satisfy 3x-2y+k=0.

3*5-2*6+k =
0.

15-12+k=0

3+k =
0

k = -3.

Put k= -3 in the equation
3x-2y+k = 0 and  we get 3x-2y-3 = 0.

So the required equation which
is perpendicular to 2x+3y+4 = 0 and passes through the point (5,6) is  3x-2y -3 =
0.