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DENIUS [597]
3 years ago
9

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
Mathematics
2 answers:
Alenkinab [10]3 years ago
6 0
<span>Triangle PQR is a right triangle. First we have to find the length of each side of the triangle. This can be done using the points provided, along with the Pythagorean theorem, which is a^2+b^2=c^2. PR^2 = (7- -2)^2+(3-5)^2 = 85 => PR = sqrt(85) QR^2 = (7- -1)^2+(3-1)^2 = 68 => QR = sqrt(68) QP^2 = (1-5)^2+(-1 - -2)^2 =17 => QP = sqrt(17) Now that we have the sides of the triangle, we can put them into the Pythagorean theorem again to see that it works out: (Sqrt(17))^2 + (sqrt(68))^2 = (sqrt(85))^2 17 + 68 = 85 85 = 85 Since the Pythagorean theorem works for right triangles, the triangle is indeed a right triangle.</span>
maw [93]3 years ago
5 0

The triangle PQR is a right angled triangle.

Further explanation:

The formula for distance between the two points can be expressed as follows,

\boxed{{\text{Distance}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}}

The Pythagorean formula can be expressed as,

\boxed{{H^2} = {P^2} + {B^2}}.

Here, H represents the hypotenuse, P represents the perpendicular and B represents the base.

Given:

The points of the triangle PQR are P \left({ - 2,5}\right), Q \left({ - 1,1}\right) and R \left({7,3}\right).

Explanation:

The distance between the points P \left({ - 2,5}\right) and Q \left( { - 1,1} \right) can be obtained as follows,

\begin{aligned}PQ&= \sqrt {{{\left({ - 1 + 2} \right)}^2} + {{\left( {1 - 5} \right)}^2}}\\&= \sqrt {{1^2} + {{\left({ - 4}\right)}^2}}\\&= \sqrt {1+16}\\&= \sqrt {17}\\\end{aligned}

The side PQ is \sqrt {17}.

The distance between the points P \left({ - 2,5}\right) and R \left({7,3}\right) can be obtained as follows,

\begin{aligned}PR&= \sqrt {{{\left( {7 + 2}\right)}^2} + {{\left( {3 - 5} \right)}^2}}\\&= \sqrt {{9^2} + {{\left({ - 2} \right)}^2}}\\&= \sqrt {81 + 4}\\&=\sqrt {85}\\\end{aligned}

The side PR is \sqrt {85}.

The distance between the points Q \left( { - 1,1} \right) and R\left( {7,3} \right) can be obtained as follows,

\begin{aligned}QR&= \sqrt {{{\left( {7 + 1}\right)}^2} + {{\left({3 - 1}\right)}^2}}\\&=\sqrt {{8^2}+{{\left({- 2} \right)}^2}}\\&= \sqrt {64 + 4}\\&= \sqrt {68}\\\end{aligned}

The side QR is \sqrt {68}.

In right angle triangle we can use the Pythagoras formula.

The square of side PQ is {\left( {\sqrt {17} }\right)^2}=17.

The square of side PR is {\left({\sqrt {85}}\right)^2}=85.

The square of side QR is {\left({\sqrt{68} }\right)^2}=68.

In triangle PQR.

\begin{aligned}P{R^2}&= P{Q^2}+Q{R^2}\\85&= 17+ 68\\85 &= 85\\\end{aligned}

The triangle PQR is a right angled triangle.

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Coordinate Geometry

Keywords: Coordinates, vertices, triangle PQR, right angle triangle, P(-2,5), Q(-1,1), R(7,3).

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