Question

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

Answer 1

The answer is yes, and I will give you the calculation soon

PQ=$\sqrt{3^{2} +1^{2} }$=$\sqrt{10}$
QR=$\sqrt{ 2^{2} + 6^{2} } =2 \sqrt{10}$
PR=$\sqrt{ 5^{2} + 5^{2} } =5 \sqrt{2}$

$\sqrt{10} ^{2} + (2\sqrt{10}) ^{2} =(5 \sqrt{2} )^{2}$

Answer 2

Remember.

A right triangle has sides a^2 + b^2 = c^2. Conversely, if a triangle has sides a^2+b^2=c^2 then it is a right triangle.

So first, find the distance between PQ, QR and RP. Then check to see if any combinations. Does PQ^2 + QR^2  = RP^2,  QR^2 +RP^2  = PQ^2 or  RP^2 + PQ^2  = QR^2. If so, you have a right triangle.