Question -
Answer -
We know that the area ofa triangle whose vertices are (x1, y1), (x2, y2), and
(x3, y3) is the absolutevalue of the determinant (Δ), where
It is given that thearea of triangle is 4 square units.
∴Δ = ± 4.
(i) The area of thetriangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,
Δ =
∴−k + 4 = ± 4
When −k + 4= − 4, k = 8.
When −k + 4= 4, k = 0.
Hence, k =0, 8.
(ii) The area of thetriangle with vertices (−2, 0), (0, 4), (0, k) is given by therelation,
Δ =
∴k − 4 = ± 4
When k −4 = − 4, k = 0.
When k −4 = 4, k = 8.
Hence, k =0, 8.