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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.
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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.
AB2 = 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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