Question

Find the coordinates of the foot of the perpendicular from (-2,6) on the line
2x+3y-1=0. �

Answer 1

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Answer on Question #75327 – Math – Analytic Geometry

Question

Find the coordinates of the foot of the perpendicular from (-2,6) on the line $2x+3y-1=0$.

Solution

If $y = kx + b$ is the equation of the perpendicular passing through (-2,6), so $6 = -2k + b$; $k = (6-b)/(-2)$

The equation $2x+3y-1=0$ given in the question is rewritten in the following form:

$$y = -2x/3 + 1/3.$$

Because the two lines are perpendicular,

$$(6-b)/(-2) = -1/(-2/3) = 3/2; \quad b = 9; \quad k = (6-b)/(-2) = (6-9)/(-2) = 3/2.$$

One gets $y = 3/2x+9$ is the equation of the perpendicular to the line $2x+3y-1=0$ passing through (-2,6).

Now we find the coordinates of the foot of the perpendicular:

$$-2x/3 + 1/3 = 3x/2 + 9;$$ $$x = -4; \quad y = -2*(-4)/3 + 1/3 = 3.$$

Answer: (-4, 3).