Question

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Answer 1

Hiee ❤️

Given

Let coordinates of triangle be A(5,-2), B(6,4), C(7,-2).

Solution

$\sf \small \green{distance \: formula = \sqrt{( x_{2} - x_{1}) {}^{2} + (y_{2} - y_{1}) {}^{2} } }$

Distance between A and B,

$\sf\longrightarrow \: AB = \sqrt{(6 - 5) {}^{2} + (4 - ( - 2)) {}^{2} } \\ \sf\longrightarrow \: AB = \sqrt{1 + 36} \\ \sf\longrightarrow \: \boxed{ \bf \: AB = \sqrt{37} }$

Distance between B and C,

$\sf\longrightarrow \: </strong><strong>BC</strong><strong> = \sqrt{(7 - 6) {}^{2} + ( - 2 - 4) {}^{2} } \\ \sf\longrightarrow \: </strong><strong>BC</strong><strong> = \sqrt{1 + 36} \\ \sf\longrightarrow \: \boxed{ \</strong><strong>b</strong><strong>f \: </strong><strong>BC</strong><strong> = \sqrt{37} }$

Distance between A and C,

$\sf\longrightarrow \: AC = \sqrt{(7 - 5) {}^{2} + ( - 2 - ( - 2)) {}^{2} } \\ \sf\longrightarrow \: AC = \sqrt{4 + 0} \\ \sf\longrightarrow \: \boxed{ \bf \: AC= 2}$

Two sides of triangle are equal, AB = BC=√37.

So yes, these are the vertices of an isosceles triangle.

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I hope it helps uh ❤️✌️

Answer 2

I hope it help you.....