Question

Show that the points (-2, 3), (8, 3) and (6, 7) are the vertices of a right triangle. (2013D)
Solution​

Answer 1

Given :

We are given three points :

• (-2,3)
• (8,3)
• (6,7)

Aim :

Here, we should show that these points are the vertices of a right angled triangle.

Explaination :

The given points are to be shown that they are the vertices of a right angled triangle. For this, firstly we shall find the distance between them individually. Then secondly, we shall use inverse of Pythagoras theorem for their distances and prove that according to the question.

Solution :

Let the points be :

• A = (-2,3)
• B = (8,3)
• C = (6,7)

Let the right angled triangle be ∆ABC wherein <C=90°.

Now, the distance are :

_______________________________________

• Distance between AB :

Here :

★ $\sf x_1:-2,~ x_2:8$

★ $\sf y_1:3,~ y_2:3$

So, the distance is :

★ $\bf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\implies \sf \sqrt{ {(8 - ( - 2))}^{2} + {(3 - 3)}^{2} }$

$\implies \sf \sqrt{ {(8 + 2)}^{2} + (0) ^{2} }$

$\sf \implies \sqrt{ {(10)}^{2} }$

$\implies \bf10units$

_______________________________________

• Distance between BC :

Here :

★ $\sf x_1:8,~x_2:6$

★ $\sf y_1:3,~ y_2:7$

So,the distance is :

★ $\bf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\implies \sf \sqrt{ {(6 - 8)}^{2} + {( 7- 3)}^{2} }$

$\sf \implies\sqrt{ {( - 2)}^{2} + {(4)}^{2} }$

$\implies \sf \sqrt{4 + 16}$

$\implies \bf \sqrt{20} units$

_______________________________________

• Distance between AB :

Here :

★ $\sf x_1:6,~x_2:-2$

★ $\sf y_1:7,~y_2:3$

So, the distance is :

★ $\bf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\sf \implies \sqrt{ {( - 2 - 6)}^{2} + {(3 - 7)}^{2} }$

$\implies \sf \sqrt{ {( - 8)}^{2} + {( - 4)}^{2} }$

$\implies \sf \sqrt{64 + 16}$

$\implies \bf \sqrt{80} units$

_______________________________________

From the Pythagoras theorem, we can say that :

• AB² + BC² = AC²

Here :

• AB = Perpendicular
• BC = Base
• AC = Hypotenuse

Now, let's substitute the distances and verify :

$\sf \implies {( \sqrt{80} )}^{2} units + {( \sqrt{20} })^{2} units = {( 10 })^{2} units$

$\sf\implies80units + 20units = 100units$

$\implies \sf100units = 100units$

Here, the Perpendicular² + Base² = Hypotenuse² is proved.

According to the inverse of Pythagoras theorem, if the square of the length of the longest side in a triangle is equal to the sum of the squares of the other two sides, the triangle is a right angled triangle.

Hence, proved!