Prove that the area of triangle whose vertices are (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent of t

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Solution

Let A = ( $x_{1}$ , $y_{1}$ ) = (t, t-2), B( $x_{2}$ , $y_{2}$ ) = (t + 2, t + 2) and C = ( $x_{3}$ , $y_{3}$ ) = (t + 3, t) be the vertices of the given triangle. Then,

Area of $Δ$ ABC = $\frac{1}{2}$ | $x_{1}$ ( $y_{2}$ - $y_{3}$ ) + $x_{2}$ ( $y_{3}$ - $y_{1}$ ) + $x_{3}$ ( $y_{1}$ - $y_{2}$ )|

Area of $Δ$ ABC = $\frac{1}{2}$ |{t(t + 2 - t) + (t + 2)(t - t + 2) + (t + 3)(t - 2 - t - 2)}|

Area of $Δ$ ABC = $\frac{1}{2}$ |{2t + 2t + 4 - 4t - 12}| = |-4| = 4 sq. units

Clearly, area of $Δ$ ABC is independent of t.

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