Question

Prove that the following points are the vertices of a right triangle.

(-4 ,-3),(-2,2) and (8, -2)

(inappropriate answers will be reported) ​

Answer 1

If three given points have to be vertices of a right triangle, then any two lines joining any two among the three points will be perpendicular to each other.

Here the points A(-4, -3), B(-2, 2) and C(8, -2) are given. If the triangle ABC has to be a right triangle, then any two among the lines AB, BC and AC should be perpendicular to each other.

Conditions for two lines being perpendicular to each other is that the product of their slopes should be -1.

Slope of line AB,

$\longrightarrow m_{AB}=\dfrac{2-(-3)}{(-2)-(-4)}$

$\longrightarrow m_{AB}=\dfrac{5}{2}$

Slope of line BC,

$\longrightarrow m_{BC}=\dfrac{2-(-2)}{(-2)-8}$

$\longrightarrow m_{BC}=-\dfrac{2}{5}$

We see the product of slopes of AB and BC is -1.

$\longrightarrow m_{AB}\times m_{BC}=\dfrac{5}{2}\times-\dfrac{2}{5}$

$\longrightarrow m_{AB}\times m_{BC}=-1$

This means AB and BC are perpendicular lines.

This implies ABC is a right triangle, right angled at B.

Hence Proved!

Answer 2

Given ,

The three points of Δ are A(-4,-3) , B(-2,2) and C(8,-2)

We know that , the distance b/w two points is given by

$\boxed{ \tt{Distance = \sqrt{ {( x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} } }}$

Thus ,

$\tt \implies AB = \sqrt{ {( - 2 + 4)}^{2} + {(2 + 3)}^{2} }$

$\tt \implies AB = \sqrt{ {(2)}^{2} + {(5)}^{2} }$

$\tt \implies AB = \sqrt{4 + 25}$

$\tt \implies AB = \sqrt{29}$

Similarly ,

$\tt \implies BC = \sqrt{ {(8 + 2)}^{2} + {( - 2 - 2)}^{2} }$

$\tt \implies BC = \sqrt{ {(10)}^{2} + {( - 4)}^{2} }$

$\tt \implies BC = \sqrt{100 + 16}$

$\tt \implies BC = \sqrt{116}$

Now ,

$\tt \implies AC = \sqrt{ {(8 + 4)}^{2} + {( - 2 + 3)}^{2} }$

$\tt \implies AC = \sqrt{ {(12)}^{2} + {(1)}^{2} }$

$\tt \implies AC = \sqrt{144 + 1}$

$\tt \implies AC = \sqrt{145}$

It is observed that ,

(AC)² = (AB)² + (BC)²

Therefore , it is a vertices of right angled triangle