Question

show that the following points taken in order from an isosceles triangle. A=(6,-4),B(-2,-4),C(2,10)

Answer 1

here is your answer

Answer 2

Answer with Step-by-step explanation:

Given that A(6,-4), B(-2,-4), C(2,10)

To prove: ABC is an isosceles triangle

As we know the distance formula: $d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Now A(6,-4),  B(-2,-4)

$AB= \sqrt{(-2-6)^2 + (-4-(-4))^2}= \sqrt{(-8)^2 + (-4+4)^2}= \sqrt{64+0} = 8 \text { unit}$

B(-2,-4), C(2,10)

$BC= \sqrt{(2-(-2))^2 + (10-(-4))^2}= \sqrt{(2+2)^2 + (10+4)^2}\\\\= \sqrt{(2+2)^2 + (10+4)^2}\\\\=\sqrt{4^2+14^2} =\sqrt{16+196} =\sqrt212} \text { unit}$

A(6,-4) , C(2,10)

$AC=\sqrt{(2-6)^2 + (10-(-4))^2}= \sqrt{(-4)^2 + (14)^2}\\\\ =\sqrt{16+196} =\sqrt212}\text { unit}$

As, AC and BC are equal so it is an isosceles triangle