Question

The coordinates of the vertices of ∆PQR are P(-2,5), Q(-1,1), and R(7,3). Determine whether ∆PQR is a right triangle. Show your work and explain your answer!

Answer 1

Given

∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

Determine whether ∆PQR is a right triangle

To proof

As given ∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

First find out the sides of triangle

FORMULA

Distance formula between two points

$D^{2}= (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

 Distance   between two points P(-2,5) and Q(-1,1)

$PR = \sqrt{(-1+2)^{2}+(1-5)^{2} }$

$PR = \sqrt{17}$

Distance between two points Q(-1,1)and  R(7,3)

$QR = \sqrt{(7+1)^{2} +(3-1)^{2} }$

$QR =\sqrt{68}$

Distance between two points  R(7,3) and P(-2,5)

$RP =\sqrt{(-2-7)^{2} + (5-3)^{2} }$

$RP=\sqrt{85}$

now show that ∆PQR is a right triangle

$RP^{2} = PQ^{2} +QR^{2}$

Putting the value given above

$(\sqrt{85}) ^{2} = \sqrt{17} ^{2} +\sqrt{68} ^{2}$

85 = 17 +68

85 =85

In the right triangle

HYPOTENUSE² = BASE² + PERPENDICULAR²

This is prove above

Hence ∆PQR is a right triangle

Hence proved