Question

Determine whether QRS is a right triangle for the given vertices. Explain.

Answer 1

Answer:

yes; QR = √200, QS = √8 , RS = √208 ; QR^2 + QS^2 = RS^2

Step-by-step explanation:

To determine the triangle as a right triangle we have to find the lengths of the sides

The distance formula will be used:

$d = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2} }\\ So,\\QR=\sqrt{(-3-7)^{2}+(0+10)^{2}}\\=\sqrt{(-10)^{2}+(10)^{2}}\\=\sqrt{100+100}\\=\sqrt{200}$

$RS=\sqrt{(9+3)^{2}+(-8-0)^{2}}\\=\sqrt{(12)^{2}+(-8)^{2}}\\=\sqrt{144+64}\\=\sqrt{208}$

$QS=\sqrt{(9-7)^{2}+(-8+10)^{2}}\\=\sqrt{(2)^{2}+(2)^{2}}\\=\sqrt{4+4}\\=\sqrt{8}$

Using Pythagoras theorem, we can see that

$RS^2 = QR^2+QS^2\\(\sqrt{208})^2=(\sqrt{200})^2 +(\sqrt{8})^2\\208 = 200+8\\208=208$

Which proves that given triangle is a right triangle ..