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Find values of k if area of triangle is 4 square units and vertices are (−2, 0), (0, 4), (0, k)

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#### Solution

Area of a triangle = `1/2 abs ((x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1))`

`=> 1/2 abs ((-2,0,1),(0,4,1),(0, k,1))`

`= 1/2 [-2 (4 - k) - 0 (0 - 0) + 1 (0 - 0)]`

`= 1/2 xx (-2) (4 - k)`

= k - 4

= `pm 4`

When, k - 4 = 4, `therefore k = 8`

k - 4 = 4 `therefore k = 0`

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