Chemistry, asked by tvithrv, 1 month ago


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

Answers

Answered by yashvihmodi2006
0

Yes. They are the vertices of an isosceles triangle. The distance (or length) between points (5,-2) and (6,4) is equal to the distance between the points (6,4) and (7,-2).

As, two sides of the triangle are equal, these points are the vertices of an isosceles triangle.

Answered by tennetiraj86
4

Explanation:

Given :-

(5,-2), (6,4) and (7,-2)

To find :-

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

Solution :-

Given points are (5,-2), (6,4) and (7,-2)

To prove that they are the vertices of an Isosceles triangle we have to prove that the lengths of any two sides are equal.

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

Distance between (5,-2), (6,4):-

Let (x1, y1) = (5,-2) => x1 = 5 and y1 = -2

Let (x2, y2) = (6,4) => x2 = 6 and y2 = 4

The distance between two points

=>√[(6-5)²+(4-(-2))²]

=> √[1²+(4+2)²]

=> √(1+6²)

=>√(1+36)

=>√37

=> The distance = √37 units ------(1)

Distance between (6,4), (7,-2):-

Let (x1, y1) = (6,4) => x1 = 6 and y1 = 4

Let (x2, y2) = (7,-2) => x2 = 7 and y2 = -2

The distance between two points

=>√[(7-6)²+(-2-4²]

=> √[1²+(-6)²]

=>√(1+36)

=> √37

=> The distance = √37 units ------(2)

Distance between (5,-2),(7,-2):-

Let (x1, y1) = (5,-2) => x1 = 5 and y1 = -2

Let (x2, y2) = (7,-2) => x2 = 7 and y2 = -2

The distance between two points

=>√[(7-5)²+(-2-(-2))²]

=> √[2²+(-2+2)²]

=> √(4+0²)

=>√(4+0)

=> √4

=> 2

=> The distance = 2 units ------(3)

From (1) ,(2)&(3)

The distance between two sides are equal.

Given points are the vertices of an Isosceles triangle.

Hence, Proved.

Answer:-

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

Used formulae:-

  • To prove that they are the vertices of an Isosceles triangle we have to prove that the lengths of any two sides are equal.

  • The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
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