Question

What is the equation of the line perpendicular to 2x-3y=13 that passes through the point (-6,5)?

Answer 1

y = (-3/2)x - 4

Explanation:

Recall that a line perpendicular to a given line has a slope which is the negative reciprocal of the given line. The given line can be re-written as -3y = -2x + 13, or

y = (2/3)x - 13/3

and so the slope of a line perpendicular to this given line is -3/2.

Then the desired equation, in point-slope form, is

y - 5 = (-3/2)(x + 6) (This is in point-slope form.)

Alternatively,

2y - 10 = -3(x + 6) = -3x - 18, or

2y = 10 - 3x - 18, or

2y = -3x - 8, or y = (-3/2)x - 4 (This is in slope-intercept form.)