# Show that the Points A(-5,6), B(3,0) and C(9,8) Are the Vertices of an Isosceles Right-angled Triangle. Calculate Its Area. - Mathematics

Show that the points A(-5,6), B(3,0) and C(9,8) are the vertices of an isosceles right-angled triangle. Calculate its area.

#### Solution

Let the given points be A(-5,6), B(3,0) and C(9,8).

AB = sqrt((3-(-5))^2 +(0-6)^2) = sqrt((8)^2 +(-6)^2 = sqrt(64+36) = sqrt(100) = 10 units

BC = sqrt((9-3)^2 +(8-0)^2) = sqrt((6)^2 +(8)^2) = sqrt(36+64) = sqrt(100) = 10 units

AC = sqrt((9-(-5))^2 +(8-6)^2) = sqrt((14)^2 +(2)^2) = sqrt(196+4) = sqrt(200) = 10sqrt(2) units

Therefore,  AB =  BC = units

Also, (AB)^2 +(BC)^2 = (10)^2 +(10)^2 = 200

and (AC)^2 =(10sqrt(2))^2 = 200

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

This show that  Δ ABC  is right angled at B. Therefore, the points A(-5,6), B(3,0) and C(9,8). are the vertices of an isosceles rightangled triangle

Also, area of a triangle = 1/2 xx"base"xx "height"

If AB is the height and BC is the base,

Area= 1/2 xx 10 xx 10

= 50 square units

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

RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 24
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