Question

$(5,-2),(6,4)$ and $(7,-2)$ are the vertices of an isosceles triangle.

• A. True
• B. False

Answer 1

The correct option is A True

Let $A(5,-2),B(6,4)$ and $C(7,-2)$ are the vertices of triangle $ABC$

Distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

Now,

$AB=\sqrt{(5-6{)}^2+(-2-4{)}^2}=\sqrt{1+36}=\sqrt{37}$

$BC=\sqrt{(6-7{)}^2+(4+2{)}^2}=\sqrt{1+36}=\sqrt{37}$

$AC=\sqrt{(5-7{)}^2+(-2+2{)}^2}=\sqrt{4}=2$

$AB=BC$

so $ABC$ is an isosceles triangle.