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. Please a breakdown of how you calculate this

Answer 1

$PQ = \sqrt{(3-0)^2+(2-1)^2}= \sqrt{9+1}= \sqrt{10} \\ QR= \sqrt{(5-3)^2+(-4-2)^2}= \sqrt{4+36}= \sqrt{40} \\PR= \sqrt{(5-0)^2+(-4-1)^2}= \sqrt{25+25}= \sqrt{50}$

PR is the longest side.
The triangle will be right if
PQ² + QR² = PR²

PQ² + QR² = (√10)² + (√40)² = 10 + 40 = 50
PR² = (√50)² = 50
50=50  so, ΔPQR is a right triangle.

Answer 2

PR is the longest side.

The triangle will be right if

PQ² + QR² = PR²

PQ² + QR² = (√10)² + (√40)² = 10 + 40 = 50

PR² = (√50)² = 50