Question

check whether (5,-2),(6,4) and(7,-2) are the vertices of an isosceles triangle​

Answer 1

$\large \pmb{\mathfrak{➜ \: Given:}}$

We are given three points (5,-2), (6,4) and (7,-2).

$\large \pmb{\mathfrak{➜ \: To \: find/To \: check:}}$

We've to check whether the given points are the vertices of an isosceles triangle or not.

$\large \pmb{\mathfrak{➜ \: Solution:}}$

Let the three points be :

• A = (5,-2)
• B = (6,4)
• C = (7,-2)

We know that, in an isosceles triangle, two sides are equal, so the either of the following conditions should be fulfilled :

• AB = AC
• AC = BC
• BC = AB

The distance of AB, AC and BC can be given by the distance formula :

$\underline{\boxed{\pmb{\rm{Distance \: formula= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} }}}}$

$\implies\pmb{\rm{Distance \: of \: AB:}}$

${\pmb{\rm {Here, \: x_1=5, \: x_2=6~;~y_1= - 2, \: y_2=4}}}$

$\implies{\pmb{\rm{ \sqrt{(6 - 5)^2+(4 - ( - 2))^2} }}}$

$\implies{\pmb{{\rm{ \sqrt{(1)^2+(4+2)^2} }}}}$

$\implies {\pmb{\rm{ \sqrt{(1)^2+(6)^2} }}}$

$\implies{\pmb{\rm{ \sqrt{1+36} }}}$

$\implies{\pmb{\rm{ \sqrt{37} }}}$

$\implies{\pmb{\rm{Distance \: of \: BC:}}}$

$\pmb{\rm{Here, \: x_1=6, \: x_2=7 \: ; \: y_1=4, \: y_2= - 2}}$

$\implies{\pmb{\rm{ \sqrt{(7 - 6)^2+( - 2 - 4)^2} }}}$

$\implies{\pmb{\rm{ \sqrt{(1)^2+(-6)^2} }}}$

$\implies{\pmb{\rm{ \sqrt{1+36} }}}$

$\implies{\pmb{\rm{ \sqrt{37} }}}$

$\implies{\pmb{\rm{Distance \: of \: AC:}}}$

$\pmb{\rm{Here, \: x_1=5 \: ,x_2 = 7\: ;\: y_1= - 2, \: y_2= - 2}}$

$\implies{\pmb{\rm{ \sqrt{(7 - 5)^2+( - 2 - ( - 2))^2} }}}$

$\implies{\pmb{\rm{ \sqrt{(2)^2+( - 2 + 2)^2} }}}$

$\implies{\pmb{\rm{ \sqrt{(2)^2+(0)^2} }}}$

$\implies{\pmb{\rm{ \sqrt{4} }}}$

$\implies{\pmb{\rm{2}}}$

Hence,

• AB = √37
• BC = √37
• AC = 2

Therefore, AB = BC (√37=√37).

Since, it fulfilled the condition AB = BC, ∆ABC is an isosceles triangle.

_________________________

Swipe your screen from left to right in order to view the answer completely!

Answer 2

Answer:

We are given three points (5,-2), (6,4) and (7,-2).

We've to check whether the given points are the vertices of an isosceles triangle or not.

Let the three points be :

A = (5,-2)

B = (6,4)

C = (7,-2)

We know that, in an isosceles triangle, two sides are equal, so the either of the following conditions should be fulfilled :

AB = AC

AC = BC

BC = AB

The distance of AB, AC and BC can be given by the distance formula :

Hence,

AB = √37

BC = √37

AC = 2

Therefore, AB = BC (√37=√37).

Since, it fulfilled the condition AB = BC, ∆ABC is an isosceles triangle.

Step-by-step explanation:

Plz mark as Brainliest...