Thread: analytic geometry triangle

A triangle has vertices P (6, 7), Q(2 -1) and R(-2, 1). How do I use the slopes of the sides to show that PQR is right angled.

Originally Posted by matty

A triangle has vertices P (6, 7), Q(2 -1) and R(-2, 1). How do I use the slopes of the sides to show that PQR is right angled.

Find the slopes of PQ, QR, and PR.

If the product of any two of the slopes is -1, then you have a right angle, thus, a right triangle.

Use the equation

y2-y1
x2-x1

to find the slopes of the lines connecting the points...

the slopes that are negative reciprocals are the 90 degree angle because it makes a perpendicular line

Two lines are perpendicular if their slopes are negative reciprocals.

The slope of the line between (-2, 1) and (2, -1) is:
(1 - (-1)) / (-2 - 2) = 2 / -4 = -1/2

The slope of one of the other lines is 2. (your job to figure out which one)

Those two lines are negative reciprocals - so they are perpendicular.
So the triangle is a right triangle.