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

FORMULA

Distance formula between two points

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

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

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

now show that ∆PQR is a right triangle

Putting the value given above

85 = 17 +68

85 =85

In the right triangle

HYPOTENUSE² = BASE² + PERPENDICULAR²

This is prove above

Hence ∆PQR is a right triangle

Hence prove n

Explanation:

Answer 2

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 = √(17)`

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

`QR = \sqrt{(7+1)^(2) +(3-1)^(2)  }`

`QR =√(68)`

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

`RP =\sqrt{(-2-7)^(2) + (5-3)^(2)  }`

`RP=√(85)`

now show that ∆PQR is a right triangle

`RP^(2) = PQ^(2) +QR^(2)`

Putting the value given above

`(√(85)) ^(2) = √(17) ^(2) +√(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