Question

Prove that the points A ( -5, 4 ), B ( -1, -2 ) and C ( 5, 2 ) are the vertices of a right angled isosceles triangle.​

Answer 1

Answer:

we have to prove that A(-5,4), B(-1, -2) and C(5,2) are the vertices of an isosceles right angled triangle.

use distance formula,

AB = \sqrt{(-5+1)^2+(4+2)^2}=\sqrt{16+36}=2\sqrt{13}

(−5+1)

2

+(4+2)

2

=

16+36

=2

13

BC = \sqrt{(-1-5)^2+(-2-2)^2}=\sqrt{36+16}=2\sqrt{13}

(−1−5)

2

+(−2−2)

2

=

36+16

=2

13

CA=\sqrt{(5+5)^2+(2-4)^2}=\sqrt{100+4}=2\sqrt{26}

(5+5)

2

+(2−4)

2

=

100+4

=2

26

here, it is clear that,

AB² + BC² = 52 + 52 = 104 = CA²

from Pythagoras theorem, we know any triangle will be a right angled triangle when sides of triangle follow above condition.

so, ABC is a right angled triangle.

Let D (x, y) such that ABCD is a square.

we know, a square is also a parallelogram.

so, midpoint of diagonal of AC =midpoint of diagonal BD

{(5 - 5)/2, (4 + 2)/2} = {(x - 1)/2, (y - 2)/2}

or, (0, 3) = {(x - 1)/2, (y - 2)/2}

(x - 1)/2 = 0 => x = 1

and (y - 2)/2 = 3 => y = 8

hence, D = (1, 8)