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
[tex]D^{2}= (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}[/tex]
Distance between two points P(-2,5) and Q(-1,1)
[tex]PR = \sqrt{(-1+2)^{2}+(1-5)^{2} }[/tex]
[tex]PR = \sqrt{17}[/tex]
Distance between two points Q(-1,1)and R(7,3)
[tex]QR = \sqrt{(7+1)^{2} +(3-1)^{2} }[/tex]
[tex]QR =\sqrt{68}[/tex]
Distance between two points R(7,3) and P(-2,5)
[tex]RP =\sqrt{(-2-7)^{2} + (5-3)^{2} }[/tex]
[tex]RP=\sqrt{85}[/tex]
now show that ∆PQR is a right triangle
[tex]RP^{2} = PQ^{2} +QR^{2}[/tex]
Putting the value given above
[tex](\sqrt{85}) ^{2} = \sqrt{17} ^{2} +\sqrt{68} ^{2}[/tex]
85 = 17 +68
85 =85
In the right triangle
HYPOTENUSE² = BASE² + PERPENDICULAR²
This is prove above
Hence ∆PQR is a right triangle
Hence proved
Answer:
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
Step-by-step explanation:
