Question

Check whether (5,-2),(6,4)and (7,-2) are the vertical of an isosceles triangle.​

Answer 1

Step-by-step explanation:

Let Say

A ( 5 , -2)

B (6 , 4)

C ( 7 , -2)

Length of each sides

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

AC =\sqrt{(7-5)^2 + (-2 -(-2))^2} = \sqrt{2^2 + 0^2} = \sqrt{4 + 0} = \sqrt{4} = 2

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

AB = BC = √37

Hence verified that these vertices are of an isosceles triangles

I think it's helpful for you

thanks #_brainly_mate

Pleazzz mark me as brainliest

And follow me I follow you back

Answer 2

Step-by-step explanation:

Let the vertical of triangle be P(5,–2),Q(6,4) and R(7,–2).

$Then, \: PQ = \sqrt{(6 - 5 {)}^{2} + (4 + 2 {)}^{2} }$

$[∵distnce = \sqrt{ (x_{2} - x_{1}) + ( y_{2} - y_{1} }) ]$

$= \sqrt{(1 {)}^{2} + (6 {)}^{2} } = \sqrt{1 + 36} = \sqrt{37} units$

$and \: RP = \sqrt{(5 - {7)}^{2} + ( - 2 + {2})^{2} }$

$= \sqrt{( - 2 {)}^{2} + 0 } = \sqrt{4} = 2 \: units$

Here, we see that PQ =QR i.e. two sides of a triangle are equal.

Hence, P, Q and R will form the vertical of an isosceles triangle.

_______________________________