Question

NEED HELP ASAP! Triangle PQR has vertices at the following coordinates: P(0, 1), Q(3, 2), and R(5, -4). Determine wether or not triangle PQR is a right triangle. Show all calculations for full credit.

Answer 1

We will apply the formula:
AB = rad((Xb-Xa)^2+(Yb-Ya)^2)

PQ = rad((3-0)^2+(2-1)^2) = rad(9+1) = rad(10)
PR = rad((5-0)^2+(-4-1)^2) = rad(25+25) = rad(50)
QR = rad((5-3)^2+(-4-2)^2) = rad(4+36)=rad(40)

We can apply the reverse of the Pythagorean Theorem(If a triangle has sides of lengths a, b, and c where c is the longest length and c^2 = a^2 + b^2, then the triangle is a right triangle)

PQ^2+QR^2=PR^2
10+40=50 TRUE

The final statement is that PQR is a right triangle with <PQR as a right angle.