Question

Check whether (5, –2), (6,4) and (7, –2) are the vertices of an isosceles triangle.
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Answer 1

Answer:

We can check the required by application of 'distance formula' which is used to find the distance between two points whose coordinates are known. If length of two sides will be equal then the triangle will be isosceles.

Let A(x₁, y₁) and B (x₂, y₂) represent 2 points in same plane, then by distance formula, the distance 'd' between these points is given by

d = √ (x₂ - x₁)² + (y₂ - y₁)²

Let the given points be written as:

A (5,-2)

B (6, 4)

C (7, -2)

AB = √(6 - 5)² + (4+2)² = √( 1 + 36) =  √37 units

BC = √(7 - 6)² + (-2-4)² = √(1 + 36) = √37 units

CA = √(7-5)² + (-2+2)² = √(4+0) = 2 units

As, AB = BC = √37 units

The given vertices form an isosceles triangle.

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Answer 2

$\huge \fcolorbox{black}{lightblue}{Solution.}$

Let A ➛ (5, –2), B ➛(6, 4) and C ➛(7, –2)

Then,

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

$= \sqrt[]{(1) {}^{2} } + (6) {}^{2} = \sqrt[]{1 + 36}$

$= \sqrt[]{37 \: }$

$BC = \sqrt[]{(7 - 6) {}^{2} + ( - 2 - 4) {}^{2} }$

$= \sqrt[]{(1) {}^{2} + ( - 6) {}^{2} }$

$= \sqrt[]{1 + 36 } = \sqrt[]{37}$

$AC = \sqrt[]{(7 - 5) {}^{2} + ( - 2 - ( - 2)) {}^{2} }$

$\sqrt[]{(2) {}^{2} + 0 {}^{2} } = \sqrt[]{4} = 2$

We can see that AB = BC ≠ AC

Therefore, ∆ ABC is an isosceles triangle.

Hence, the points (5, –2), (6, 4) and (7, –2) are

the vertices of an isosceles triangle.