haydenirwin
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GEOMETRY!!!!!!! PLEASE HELP

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.

Answer :

kenzzieee
Let r = the distance from P(-2,-4) to Q(-4,-2); 
r = √{(-4 - -2)² + (-2 - -4)²} 
r = √{-2² + 2²} 
r = √{4 + 4} 
r = √8 
Let q = the distance from P(-2,-4) to R(7,-1): 
q = √{(7 - -2)² + (-1 - -4)²} 
q = √{9² + 3²} 
q = √{81 + 9} 
q = √90 
Let p = the distance from Q(-4,-2) to R(7,-1): 
p = √{(7 - -4)² + (-1 - -2)²} 
p = √{11² + 1²} 
p = √{121 + 1) 
p = √122 
If PQR is a right triangle, then: 
p² = q² + r² 
122 = 90 + 8 
122 ≠ 98 
Therefore, it is not a right triangle.

Answer:  YES, the triangle PQR is a right-angled triangle.

Step-by-step explanation:  Given that the co-ordinates of the vertices of triangle PQR are P(-2, 5), Q(-1, 1), and R(7, 3).

We are to determine whether the triangle PQR is right-angled or not.

We know that

to be a right-angled triangle, the square of the length of the largest side (hypotenuse) must be equal to the sum of the squares of the lengths of the other two sides (legs).

Now, the lengths of the sides of the triangle PQR can be calculated using distance formula as follows :

[tex]PQ=\sqrt{(-1+2)^2+(1-5)^2}=\sqrt{1+16}=\sqrt{17}~\textup{units},\\\\QR=\sqrt{(7+1)^2+(3-1)^2}=\sqrt{64+4}=\sqrt{68}~\textup{units},\\\\RP=\sqrt{(-2-7)^2+(5-3)^2}=\sqrt{81+4}=\sqrt{85}~\textup{units}.[/tex]

So, we have

[tex]PQ^2=17,~~QR=68,~~RP^2=85\\\\\Rightarrow PQ^2+QR^2=17+68=85=RP^2.[/tex]

Thus, triangle PQR is a right-angled triangle.

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