Question

question 4 of exercise 7.1 class 10 ​

Answer 1

Given :-

Point A = (5,-2)

Point B = (6,4)

Point C = (7,-2)

Two sides of any isosceles triangle are equal.

To Find :-

Check whether given points are vertices of an isosceles triangle.

Analysis :-

To check whether given points are vertices of an isosceles triangle, first find the distance between all the points.

Solution :-

Given that,

Point A = (5,-2)

Point B = (6,4)

Point C = (7,-2)

According to the question,

Since two sides of any isosceles triangle are equal. To check whether given points are vertices of an isosceles triangle, we will find the distance between all the points.

Let the points (5, – 2), (6, 4), and (7, – 2) are representing the vertices A, B, and C respectively.

$\sf AB=\sqrt{(6-5)^{2}+(4+2)^{2}} =\sqrt{(-1)^{2}+(6)^{2}} =\sqrt{37}$

$\sf BC= \sqrt{(7-6)^{2}+(-2-4)^{2}} =\sqrt{(-1)^{2}+(6)^{2}} =\sqrt{37}$

$\sf CA=\sqrt{(7-5)^{2}+(-2+2)^{2}} =\sqrt{(-2)^{2}+(0)^{2}} =2$

Here, $\sf AB=BC=\sqrt{37}$

This shows that, whether given points are vertices of an isosceles triangle.

Answer 2

Answer:

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