# Show that the Points a (7, 10), B(-2, 5) and C(3, -4) Are the Vertices of an Isosceles Right Triangle. - Mathematics

prove  that the points A (7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right triangle.

#### Solution

The given points are A (7, 10), B(-2, 5) and C(3, -4).

AB= sqrt((-2-7)^2 +(5-10)^2) = sqrt((-9)^2 +(-5)^2) = sqrt((81+25)) = sqrt(106)

BC = sqrt((3-(-2))^2 +(-4-5)^2) = sqrt((5)^2 +(-9)^2 )= sqrt((25+81) )= sqrt(106)

AC = sqrt((3-7)^2 +(-4-10)^2) = sqrt(( -4)^2 +(-14)^2) = sqrt(16+196) = sqrt(212)

Since, AB and BC are equal, they form the vertices of an isosceles triangle

Also,(AB)^2 + (BC)^2 = ( sqrt(106))^2 +( sqrt(106)^2) = 212

and (AC)^2 = (sqrt(212))^2 = 212.

Thus , (AB)^2 + (BC)^2 = (AC)^2`

This show that  ΔABC is right- angled at B. Therefore, the pointsA (7, 10), B(-2, 5) and C(3, -4). are the vertices of an isosceles rightangled triangle.

Concept: Area of a Triangle
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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.2 | Q 45 | Page 17
RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 19