Question

Show that the triangle formed by the points A(5,-1),B(-3,2) and C(1,6) is an isosceles triangle

Answer 1

Distance between AB=
$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$\sqrt{(-3-5)^2+(2+1)^2}$
$\sqrt{(-8)^2+(3)^2}$
$\sqrt{64+9}$
$\sqrt{73}$ units

Distance between BC= 
$\sqrt{(1+3)^2+(6-2)^2}$
$\sqrt{4^2+4^2}$
$\sqrt{16+16}$
$\sqrt{32}$
$\sqrt{16X2}$
$4 \sqrt{2}$ units

Distance between CA=

$\sqrt{(5-1)^2+(-1-6)^2}$
$\sqrt{4^2+(-7)^2}$
$\sqrt{16+49}$
$\sqrt{65}$ units

In an isosceles triangle, any two sides are equal. The distance of any two sides is not equal here. So, the given points do not form an isosceles triangle.

Answer 2

a(5,-1)
b(-3,2)
c(1,6)
distance²=(x2-x)²+(y²-y)²
ab²=(5--3)²+(-1-2)²
ab²=64+9=73
ab=√73
bc²=(-3-1)²+(2-6)²
=16+16
bc=√32
ac²=(5-1)²+(-1-6)²=16+49
ac=√65

there aren't any same sides...so it is not a isosceles triangle