Question:
The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be
(A) 9
(B) 3
(C) – 9
(D) 6
Solution:
Option (B) 3
We know that, the area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by
Area of triangle with vertices $(-3,0),(3,0)$ and $(0, k)$ is
$[-3(-k)-0+1(3 k)]=\pm 18$
$6 k=\pm 18$
Thus,
$k=\pm \frac{18}{6}=\pm 3$
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