Let the given points be \(A(-5,6), B(3,0) \text { and } C(9,8) \text {. }\)
\(A B =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
\(=\sqrt{(3-(-5))^2+(0-6)^2} \)
\(=\sqrt{(3+5)^2+(-6)^2}\)
\(=\sqrt{(8)^2+(-6)^2}\)
\(=\sqrt{64+36} \)
\(=\sqrt{100} \)
\(=10 \text { units }\)
\(BC =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
\( =\sqrt{(9-3)^2+(8-0)^2} \)
\( =\sqrt{(6)^2+(8)^2}\)
\(=\sqrt{36+64}\)
\(=\sqrt{100}\)
\(=10 \text { units }\)
\(A C =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
\(=\sqrt{(9-(-5))^2+(8-6)^2}\)
\(=\sqrt{(9+5)^2+(8-6)^2} \)
\( =\sqrt{(14)^2+(2)^2} \)
\(=\sqrt{196+4} \)
\(=\sqrt{200}\)
\(=10 \sqrt{2} \text { units }\)
Therefore, AB = BC
The points \(A(-5,6), B(3,0)\) and \(C(9,8)\) are the vertices of an isosceles triangle.
