Question

show that the piont taken in onder form an isosceles triangle A(2,5) ,B(2,0),C(-2,3)​

Answer 1

Solution :-

The points are :-

A ( 2 , 5 )

B ( 2 , 0 )

C ( -2 , 3 )

$\boxed{\bf Distance \ formula = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }$

$\bullet \sf \ AB = \sqrt{(2-2)^2+(5-0)^2}$

       $= \sf \sqrt{0+5^2} \\\\= \sqrt{25} \\\\= 5 \ units$

$\bullet \sf \ BC = \sqrt{(2-(-2))^2+(3-0)^2}$

        $= \sf \sqrt{4^2+3^2} \\\\= \sqrt{16+9} \\\\= \sqrt{25} \\\\= 5 \ units$

$\bullet \sf \ AC = \sqrt{((2-(-2))^2+(5-3)^2}$

       $= \sf \sqrt{4^2+2^2} \\\\= \sqrt{16+4} \\\\= \sqrt{20} \\\\= \sqrt{4 \times5} \\\\= 2 \sqrt{5} \ units$

AB = BC

That is, two sides are equal.

In an isosceles triangle two sides are equal.

So the given points are the vertices of an isosceles triangle.

Answer 2

Step-by-step explanation:

The points are :-

A ( 2 , 5 )

B ( 2 , 0 )

C ( -2 , 3 )

$\boxed{\bf Distance \ formula = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }$

$\bullet \sf \ AB = \sqrt{(2-2)^2+(5-0)^2}∙ AB= </p><p>(2−2) </p><p>2</p><p> +(5−0) </p><p>2</p><p>$

$\begin{gathered}= \sf \sqrt{0+5^2} \\\\= \sqrt{25} \\\\= 5 \ units\end{gathered} </p><p>$

$\bullet \sf \ BC = \sqrt{(2-(-2))^2+(3-0)^2}∙ BC= </p><p>(2−(−2)) </p><p>2</p><p> +(3−0) </p><p>2$

$\begin{gathered}= \sf \sqrt{4^2+3^2} \\\\= \sqrt{16+9} \\\\= \sqrt{25} \\\\= 5 \ units\end{gathered} </p><p></p><p>$

$\bullet \sf \ AC = \sqrt{((2-(-2))^2+(5-3)^2}∙$

$\begin{gathered}= \sf \sqrt{4^2+2^2} \\\\= \sqrt{16+4} \\\\= \sqrt{20} \\\\= \sqrt{4 \times5} \\\\= 2 \sqrt{5} \ units\end{gathered}$

AB = BC

That is, two sides are equal.

In an isosceles triangle two sides are equal.

So the given points are the vertices of an isosceles triangle.