What is the standard equation of the line perpendicular to line segment PQ at point Q. P(1,3) Q(2,4)?

Question

Can anyone please explain how to do this type of problem?

I know the slope would be 5/2 because the slope is perpendicular, but the fact that it is a fraction confuses me.

Answer 1

The slope is (4-3)/(2-1) = 1/1 = 1, so the perpendiculars slope is -1

So, then angle PQ/x axis is 45o. And the perpendiculars angle will be -45o

The equation of the straight line (a non vertical one) is:

y - y0 = m (x - x0)

Just substitute x0 and y0 for Qs coordinates

Ana

Answer 2

The slope of the line segment PQ is 1, so the perpendicular line has a slope of -1.

Using y = mx+b replace the slope with -1 and use point Q for your x and y values. Then you will be able to solve for b.

By the way if your answer has to be in standard form, that form is Ax + By = C. So make sure your answer is in the right format.

Hope this helps. Good luck.

Answer 3

Slope between PQ
= (4 - 3) / (2 - 1) = 1

Slope of perpendicular = negative reciprocal = -1/1

use point slope formula: (y -y1) = m(x-x1)
y - 4 = -1 (x - 2)
y - 4 = -x + 2
y = -x + 6

where did you get 5/2?