Question

If the points (4,4) , (3, 5) and (-1, -1) are the vertices of a traingle then prove that the points are the vertices of a right angled triangle.​

Answer 1

$\Large\sf{\underline{\underline{Question:-}}}$

If the points (4, 4) , (3, 5) and (-1, -1) are the vertices of a triangle then prove that the points are the vertices of a right angled triangle.

$\Large\sf{\underline{\underline{Solution:-}}}$

Let A(4,4) , B(3, 5) and C(-1, -1) be the vertices of a triangle.

Then,

slope of AB = $\large\sf\dfrac{5 - 4}{3 - 4} = \frac{1}{-1} = -1$

slope of BC = $\large\sf\dfrac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2}$

slope of CA = $\large\sf\dfrac{-1 - 4}{-1 - 4} = \frac{-5}{-5} = 1$

So, slope of AB × slope of CA = (-1) × 1 = -1

AB perpendicular to CA

$\angle\:A$ = 90°

Hence, the given three points are the vertices of a right angled triangle.

$\pink{Hope \: it \: helps}$

Answer 2

Answer:

slope of AB =$\large\sf\dfrac{5 - 4}{3 - 4} = \frac{1}{-1} = -1$

slope of BC = $\large\sf\dfrac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2}$

slope of CA = $\large\sf\dfrac{-1 - 4}{-1 - 4} = \frac{-5}{-5} = 1$

So, slope of AB × slope of CA = (-1) × 1 = -1

AB perpendicular to CA

∠A = 90°