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Prove that the Points (0, 0), (5, 5) and (-5, 5) Are the Vertices of a Right Isosceles Triangle. - Mathematics

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Question

Prove that the points (0, 0), (5, 5) and (-5, 5) are the vertices of a right isosceles triangle.

Solution

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.

By Pythagoras Theorem in a right-angled triangle, the square of the longest side will be equal to the sum of squares of the other two sides.

Here the three points are A(0,0), B(5,5) and C(5,5).

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

`AB = sqrt((0 - 5)^2 + (0 - 5)^2)`

`= sqrt((-5)^2 + (0 - 5)^2)`

`= sqrt(25 + 25)`

`AB = 5sqrt2`

`BC = sqrt((5 + 5)^2 + (5 - 5)^2)`

`= sqrt((10)^2 + (0)^2)`

`= sqrt(100)`

BC = 10

`AC = sqrt((0 + 5)^2 + (0 - 5)^2)`

`= sqrt((5)^2 + (-5)^2)`

`=sqrt(25 + 25)`

`AC =  5sqrt2`

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

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Chapter 6: Co-Ordinate Geometry - Exercise 6.2 [Page 16]

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RD Sharma Mathematics [English] Class 10
Chapter 6 Co-Ordinate Geometry
Exercise 6.2 | Q 32 | Page 16

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