Question

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

Answer 1

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

AB =$\surd (-2-7{)}^2+(5-10{)}^2$

= $\surd (-9{)}^2+(-5{)}^2$

=$\surd 81+25$=√106 units

BC = $\surd (3-(-2){)}^2+(-4-5{)}^2$

=$\surd (5{)}^2+(-9{)}^2$

= $\surd 25+81$ =√106 units

AC =$\surd (3-7{)}^2+(-4-10{)}^2$

=$\surd (-4)2+(-14)2$

= $\surd 16+196$ =$\surd 212units$

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

Also,$(AB{)}^2+(BC{)}^2$
= $\surd (106{)}^2+\surd (106{)}^2$ = √212

and $(AC{)}^2$ = $(\surd 212{)}^2$= 212

Thus, $(AB{)}^2$ +$(BC{)}^2$ =$(AC{)}^2$

This shows that ΔABC is right angled at B

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