Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle.
Answers
Given : Points (7, 10), (-2, 5) and (3, -4 ) are vertices of a triangle.
Solution :
Let A(7, 10), B (- 2, 5) and C (3, - 4 )
By using distance formula : √(x2 - x1)² + (y2 - y1)²
Vertices : A(7, 10), B (- 2, 5)
Length of side AB = √(- 2 - 7)² + (5 - 10)²
AB = √(-9)² + (-5)²
AB = √81 + 25
AB = √106 units
Vertices : B (- 2, 5) and C (3,- 4)
Length of side BC = √(3 + 2)² + ( - 4 - 5)²
BC = √(5)² + (-9)²
BC = √25 + 81
BC = √106 units
Vertices : A(7, 10),C (3, - 4 )
Length of side AC = √(3 - 7)² + (- 4 - 10)²
AC = √(-4)² + (-14)²
AC = √16 + 196
AC = √212 units
Since the 2 sides AB = BC = √106.
Therefore ∆ is an isosceles.
Now, in ∆ABC, by using Pythagoras theorem
AC² = AB² + BC²
(√212)² = (√106)² + (√105)²
212 = 106 + 106
212 = 212
Since AC² = AB² + BC²
Hence, the given vertices of a triangle is an isosceles right triangle.
Some more questions :
Prove that the points (3, 0), (6, 4) and (- 1, 3) are vertices of a right-angled isosceles triangle.
https://brainly.in/question/15937740
Prove that the points (0, 0), (5, 5) and (−5, 5) are the vertices of a right isosceles triangle.
https://brainly.in/question/15937969
Step-by-step explanation:
80 + 605 = 685 = BC^{2}
it is proved, the points (1,5).(-7, 9) and (-10, -17) are the vertices of a
right angled triangle.