Question

Show that A(6,4)B(5,_2)C(7,_2) are the vertices of an isosceles ?

Answer 1

ANSWER:

the vertices of triangle is A(6,4),B(5,-2),C(7,-2)

Using distance formula:

√(x2-x1)^2+(y2-y1)^2

AB=√(5-6)^2+(-2-4)^2

=√1^2+(-6)^2

=√1+36

=√37 units

AC=√(7-6)^2+(-2-4)^2

=√1^2+(-6)^2

=√1+36

=√37 units

therefore AB=AC

therefore ∆ABC is an isosceles triangle

Hope this help you..

Answer 2

hey mate

here is ur answer

$\huge \blue{ \boxed { \boxed{ \mathbb{ \mid \ulcorner ANSWER:- \urcorner \mid }}}}$

given points are

A(6,4)B(5,2)C(7,2)

${\boxed{\underline{Distance \: between \: two \: points=\sqrt{(x2-x1)^2+(y2-y1)^2}}}}}$

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

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

$AC=\sqrt{(7-6)^2+(2-4)^2}=\sqrt{5}units$

two sides are equal

hence

the given points form isoscels triangle

hope the answer helps

$\Huge\orange{\boxed{\boxed{\underline{THANK YOU}}}}$