Respuesta :
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.
The triangle PQR is a right angled triangle.
Further explanation:
The formula for distance between the two points can be expressed as follows,
[tex]\boxed{{\text{Distance}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}}[/tex]
The Pythagorean formula can be expressed as,
[tex]\boxed{{H^2} = {P^2} + {B^2}}.[/tex]
Here, H represents the hypotenuse, P represents the perpendicular and B represents the base.
Given:
The points of the triangle PQR are P [tex]\left({ - 2,5}\right)[/tex], Q [tex]\left({ - 1,1}\right)[/tex] and R [tex]\left({7,3}\right).[/tex]
Explanation:
The distance between the points P [tex]\left({ - 2,5}\right)[/tex] and Q [tex]\left( { - 1,1} \right)[/tex] can be obtained as follows,
[tex]\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}[/tex]
The side PQ is [tex]\sqrt {17}.[/tex]
The distance between the points P [tex]\left({ - 2,5}\right)[/tex] and R [tex]\left({7,3}\right)[/tex] can be obtained as follows,
[tex]\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}[/tex]
The side PR is [tex]\sqrt {85}.[/tex]
The distance between the points Q [tex]\left( { - 1,1} \right)[/tex] and [tex]R\left( {7,3} \right)[/tex] can be obtained as follows,
[tex]\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}[/tex]
The side QR is [tex]\sqrt {68}.[/tex]
In right angle triangle we can use the Pythagoras formula.
The square of side PQ is [tex]{\left( {\sqrt {17} }\right)^2}=17.[/tex]
The square of side PR is [tex]{\left({\sqrt {85}}\right)^2}=85.[/tex]
The square of side QR is [tex]{\left({\sqrt{68} }\right)^2}=68.[/tex]
In triangle PQR.
[tex]\begin{aligned}P{R^2}&= P{Q^2}+Q{R^2}\\85&= 17+ 68\\85 &= 85\\\end{aligned}[/tex]
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 https://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).
