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Question
Show that the points (2, 0), (–2, 0) and (0, 2) are vertices of a triangle. State the type of triangle with reason.
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Solution
Let the points be P(2, 0), Q(–2, 0) and R(0, 2).
Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
By distance formula,
d(P, Q) = `sqrt([(-2) - 2]^2 + (0 - 0)^2`
= `sqrt((-4)^2 + (0)^2`
= `sqrt(16 + 0)`
= 4 ...(i)
d(Q, R) = `sqrt([0 - (-2)]^2 + (2 - 0)^2`
= `sqrt((2)^2 + (2)^2`
= `sqrt(4 + 4)`
= `sqrt(8)` ...(ii)
d(P, R) = `sqrt((0 -2)^2 + (2 - 0)^2`
= `sqrt((- 2)^2 + (2)^2`
= `sqrt(4 + 4)`
= `sqrt(8)` ...(iii)
On adding (ii) and (iii),
d(P, Q) + d(Q, R) = `4 + sqrt(8)`
`4 + sqrt(8) > sqrt(8)`
∴ d(P, Q) + d(Q, R) > d(P, R)
∴ Points P, Q, R are non colinear points.
We can construct a triangle through 3 non collinear points.
∴ The segment joining the given points form a triangle.
Since P(Q, R) = P(P, R)
∴ ∆PQR is an isosceles triangle.
∴ The segment joining the points (2, 0), (–2, 0) and (0, 2) will form an isosceles triangle.
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