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

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We have to find the perpendicular distance of the point (5, 6) from the line -2x + 3y + 4 = 0. Now, we know that the perpendicular distance of the point (m, n) from the line Ax + By + C = 0 is given by | Am + Bn + C| / sqrt (A^2 + B^2).

Here we have A = -2, B = 3, C = 4, m = 5 and n = 6. Substituting the values in the relation we get:

D = | -2*5 + 3*6 + 4| / sqrt (-2^2 + 3^2)

=> D = | -10 + 18 + 4 | / sqrt (4 + 9)

=> D = 12 / sqrt 13

**Therefore the perpendicular distance from the point (5, 6) to the line -2x + 3y + 4 = 0 is 12 / sqrt 13.**

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.

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