Question

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

Answer 1

Step-by-step explanation:

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Answer 2

Given :-

• The given points are (5,-2), (6,4), and (7,-2)

To Find :-

• Check whether the points are vertices of an isosceles triangle ?

Formula Used :-

♣️ Distance formula :

$\bigstar \: \: \boxed{ \bf \pmb{ \: d \: = \: \sqrt{ \bigg(x_{2} - x_{1} \bigg)^{2} + \bigg (y_{2} - y_{1} \bigg)^{2} } } \: } \: \: \bigstar$

Solution :-

Given that,

$⟶\;$The given points are (5,-2), (6,4), and (7,-2)

Let,

• (5,-2) = A
• (6,4) = B
• (7,-2) = C

Firstly, We have to find the distance between A and B :

We know that,

• $\sf \pmb{x_{1}}\;$ = 5
• $\sf \pmb{x_{2}}\;$ = 6
• $\sf \pmb{y_{1}}\;$ = -2
• $\sf \pmb{y_{2}}\;$ = 4

Now,

$\sf \: \implies \: \pmb{ \: AB \: = \: \sqrt{( x_{2} - x_{1} )^{2} +( y_{2} - y_{1} )^{2} } }$

$\sf \: \implies \: \pmb{ \: AB \: = \: \sqrt{( 6 - 5 )^{2} +( 4- ( - 2) )^{2} } }$

$\sf \: \implies \: \pmb{ \: AB \: = \: \sqrt{( 1 )^{2} +( 4 + 2 )^{2} } }$

$\sf \: \implies \: \pmb{ \: AB \: = \: \sqrt{( 1 )^{2} +( 6)^{2} } }$

$\sf \: \implies \: \pmb{ \: AB \: = \: \sqrt{1 +36} }$

$\sf \: \implies \: \pmb{ \: \red{ AB \: = \: \sqrt{37} }}$

Now, We have to find the distance between B and C :

We know that,

• $\sf \pmb{x_{1}}\;$ = 6
• $\sf \pmb{x_{2}}\;$ = 7
• $\sf \pmb{y_{1}}\;$ = 4
• $\sf \pmb{y_{2}}\;$ = -2

Now,

$\sf \: \implies \: \pmb{ \: B C\: = \: \sqrt{( x_{2} - x_{1} )^{2} +( y_{2} - y_{1} )^{2} } }$

$\sf \: \implies \: \pmb{ \: B C\: = \: \sqrt{( 7 - 6 )^{2} +( - 2- ( + 4) )^{2} } }$

$\sf \: \implies \: \pmb{ \: B C\: = \: \sqrt{( 1)^{2} +( - 2- 4)^{2} } }$

$\sf \: \implies \: \pmb{ \: B C\: = \: \sqrt{1+( -6)^{2} } }$

$\sf \: \implies \: \pmb{ \: B C\: = \: \sqrt{1 +36 } }$

$\sf \: \implies \: \pmb{ \red{\: B C\: = \: \sqrt{37 } } }$

Then, We have to find the distance between C and A :

We know that,

• $\sf \pmb{x_{1}}\;$ = 7
• $\sf \pmb{x_{2}}\;$ = 5
• $\sf \pmb{y_{1}}\;$ = -2
• $\sf \pmb{y_{2}}\;$ = -2

Now,

$\sf \: \implies \: \pmb{ \: CA\: = \: \sqrt{( x_{2} - x_{1} )^{2} +( y_{2} - y_{1} )^{2} } }$

$\sf \: \implies \: \pmb{ \: CA\: = \: \sqrt{( 5 - 7 )^{2} +( - 2 - ( - 2))^{2} } }$

$\sf \: \implies \: \pmb{ \: CA\: = \: \sqrt{( - 2)^{2} +( - 2 + 2)^{2} } }$

$\sf \: \implies \: \pmb{ \: CA\: = \: \sqrt{ 4 +( 0)^{2} } }$

$\sf \: \implies \: \pmb{ \: CA\: = \: \sqrt{ 4 + 0} }$

$\sf \: \implies \: \pmb{ \: CA\: = \: \sqrt{ 4} }$

$\sf \: \implies \: \pmb{ \: CA\: = \: \sqrt{ {2} ^{2} \:} }$

$\sf \: \implies \: \pmb{ \: \red{CA\: = \: 2}}$

Hence,

∴ We proved that the points are vertices of an isosceles triangle.

$\:$

NOTE :

In isosceles triangle, two sides are equal and the other side is different in measures.