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!

It is a right triangle. Here is a picture to show why. You can also use the pythagorean theorem. The formula is c= √a^2+b^2. This can also help, to prove that this is a right triangle.

Answer Link

calculista calculista · Answer 2 · 2018-11-03T20:15:47+01:00

Step 1

Plot the vertices of triangle PQR

[tex]P(-2,5)\ Q(-1,1)\ R(7,3)[/tex]

using a graphing tool

see the attached figure

we know that the distance between two points is equal to

[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]

Step 2

Find the distance PR

[tex]P(-2,5)\ R(7,3)[/tex]

substitute the values

[tex]d=\sqrt{(3-5)^{2}+(7+2)^{2}}[/tex]

[tex]d=\sqrt{(-2)^{2}+(9)^{2}}[/tex]

[tex]dPR=\sqrt{85}\ units[/tex]

Step 3

Find the distance QP

[tex]Q(-1,1)\ P(-2,5)[/tex]

substitute the values

[tex]d=\sqrt{(5-1)^{2}+(-2+1)^{2}}[/tex]

[tex]d=\sqrt{(4)^{2}+(-1)^{2}}[/tex]

[tex]dQP=\sqrt{17}\ units[/tex]

Step 4

Find the distance QR

[tex]Q(-1,1)\ R(7,3)[/tex]

substitute the values

[tex]d=\sqrt{(3-1)^{2}+(7+1)^{2}}[/tex]

[tex]d=\sqrt{(2)^{2}+(8)^{2}}[/tex]

[tex]dQR=\sqrt{68}\ units[/tex]

Step 5

If triang;le PQR is a right triangle

then

Applying the Pythagorean Theorem

[tex]PR^{2} =QP^{2} +QR^{2}[/tex]

substitute the values

[tex]\sqrt{85}^{2} =\sqrt{17}^{2} +\sqrt{68}^{2}[/tex]

[tex]85=17+68[/tex]

[tex]85=85[/tex] ---------> is true

therefore

The answer is

The triangle PQR is a right triangle