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Maharashtra State BoardSSC (English Medium) 10th Standard

Show that the points (2, 0), (–2, 0), and (0, 2) are the vertices of a triangle. Also, a state with the reason for the type of triangle.

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Question

Show that the points (2, 0), (–2, 0), and (0, 2) are the vertices of a triangle. Also, a state with the reason for the type of triangle.

Sum
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Solution

A ≡ (2, 0) ≡ (x1, y1)

B ≡ (–2 , 0) ≡ (x2, y2)

C ≡ (0, 2) ≡ (x3, y3)

∴ ABC form a triangle.

AB = `sqrt ((x_2 - x_1)^2 + ("y"_2 - "y"_1)^2)`

= `sqrt ((-2-2)^2 + (0 - 0)^2)`

= `sqrt ((-4)^2 + 0)`

= `sqrt(16)`

= 4 units

AC = `sqrt ((x_3 - x_1)^2 + ("y"_3 - "y"_1)^2)`

= `sqrt ((0-2)^2 + (2 - 0)^2)`

= `sqrt ((-2)^2 + (2)^2)`

= `sqrt (4 + 4)`

= `sqrt(8)`

= `2sqrt(2)` units

BC = `sqrt ((x_3 - x_2)^2 + ("y"_3 - "y"_2)^2)`

= `sqrt ((0-(-2)^2) + (2 - 0)^2)`

= `sqrt ((0 + 2)^2 + (2 - 0)^2)`

= `sqrt ((2)^2 + (2)^2)`

= `sqrt(8)`

= `2sqrt(2)` units 

So, if side AC and side BC are equal then the triangle is an isosceles triangle.

AB= BC2 + AC2

`(4)^2 = (2sqrt2)^2 + (2sqrt(2))^2`

16 = 8 + 8

16 = 16

So, it is a right-angle isosceles triangle.

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2018-2019 (March) Set 1

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