# Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle. - Geometry

Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle.

#### Solution 1

Let A(3, 0), B(6, 4) and C(–1, 3) be the given points

Now,

AB= sqrt((6-3)^2+(4-0)^2)=sqrt(3^2+4^2)=sqrt(9+6)=sqrt25

BC= sqrt((-1-6)^2+(3-4)^2)=sqrt((-7)^2+(-1)^2)=sqrt(49+1)=sqrt50

AC= sqrt((-1-3)^2+(3-0)^2)=sqrt((-4)^2+3^2)=sqrt(16+9)=sqrt25

∴ AB = AC

AB2 =(sqrt25)=25

BC2= (sqrt50)=50

AC2= (sqrt25)=25

∴ AB2 + AC2 =  BC2

Thus, ΔABC is a right-angled isosceles triangle.

#### Solution 2

The distance d between two points (x_1,y_1) and (x_2, y_2) is given by the formula

d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)

In an isosceles triangle there are two sides which are equal in length.

Here the three points are A(3, 0), B(64) and C(13).

Let us check the length of the three sides of the triangle.

AB = sqrt((3 - 6)^2 + (0 - 4)^2)

= sqrt((-3)^2 + (-4)^2)

= sqrt(9 + 16)

AB = sqrt(25)

BC = sqrt((6 + 1)^2 + (4 - 3)^2)

= sqrt((7)^2 + (1)^2)

= sqrt(49 + 1)

BC = sqrt50

AC = sqrt((3 + 1)^2 + (0 - 3)^2)

= sqrt((4)^2 + (-3)^2)

= sqrt(16 - 9)

AC = sqrt25

Here, we see that two sides of the triangle are equal. So the triangle formed should be an isosceles triangle.

We can also observe that BC^2 = AC^2 + AB^2`

Hence proved that the triangle formed by the three given points is an isosceles triangle.

Concept: Coordinate Geometry
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.2 | Q 9 | Page 15