Question

Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

Answer 1

Answer:

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Step-by-step explanation:

Let ABC be the triangle where A(5,−2),B(6,4),C(7,−2)

$AB = \sqrt{(6 - 5) {}^{2} + (4 + 2) {}^{2} } = \sqrt{1 + 36} = \sqrt{37} \\ \\ BC = \sqrt{(7 - 6) {}^{2} + ( - 2 - 4) {}^{2} } = \sqrt{1 + 36} = \sqrt{37} \\ \\ AC = \sqrt{(7 - 5) {}^{2} + ( - 2 + 2) {}^{2} } = \sqrt{4 + 0} = 2$

Here AB = BC

Hence (5,−2),(6,4) & (7,−2) are vertices of isosceles triangle

Answer 2

Answer: yes it is an isosceles triangle

Step-by-step explanation:

Let the vertices be A(5,-2); B(6,4) and C(7,-2)

For the triangle to be an isosceles triangle, 2 of the sides should be equal to each other in magnitude

This implies that either AB = BC (or) BC = AC (or) AB = AC

We first need to find the length of the sides

Formula for finding the length of the side is

$\sqrt{ (x2 - x1)^{2} + (y2 - y1)^{2} }$

Where (x1, y1) and (x2, y2) are the points between which the distance is to be found

Substituting values of the vertices and solving, we get

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

=> AB = $\sqrt{ 1^{2} +6^{2} } = \sqrt{1+36} = \sqrt{37}$

=> AB = $\sqrt{37}$

Similarly while solving by substituting, we get

BC = $\sqrt{37}$

And

AC = $\sqrt{2^{2}} = 2$

Therefore, AB = BC

Thus ABC is an isosceles triangle with AB = BC