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Question
Prove that the area of a triangle with vertices (t, t −2), (t + 2, t + 2) and (t + 3, t) is independent of t.
Solution 1
Let A(t, t − 2), B(t + 2, t + 2) and C(t + 2, t) be the vertices of the given triangle.
We know that the area of the triangle having vertices (x1, y1), (x2, y2) and (x3, y3) is
`1/2|x_1(y_2-y_3)+x^2(y_3-y_1)+x_3(y_1-y_2)||`
∴ Area of ∆ABC = `|1/2[x_1(y_2-y_3)+x^2(y_3-y_1)+x_3(y_1-y_2)||`
`=|1/2[t(t+2−t)+(t+2)(t−t+2)+(t+3)(t−2−t−2)]|`
`|1/2(2t+2t+4−4t−12)|`
`|-4|`
=4 square units
Hence, the area of the triangle with given vertices is independent of t.
Solution 2
`therefore `
`rArr `
`rArr`
`rArr`
⇒ Area of ∆ABC = 4 sq. units
Hence, Area of ∆ABC is independent of t.
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