If the area of a triangle whose vertices are \((-1, 3), (1, -5)\) and \((k,2)\) where \(k \gt 0\) is 30 sq. units, then the value of \(k\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{27}{4}\)
The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is
\( \displaystyle \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right| \)
With the vertices (−1, 3), (1, −5), (k, 2) and area 30:
\( \displaystyle 30 = \frac{1}{2}\left|(-1)(-5 - 2) + 1(2 - 3) + k(3 + 5)\right| \)
\( \displaystyle 30 = \frac{1}{2}\left|7 - 1 + 8k\right| \)
\(|8k + 6| = 60\)
So
\( \displaystyle 8k + 6 = 60,\quad k = \frac{27}{4} \)
or
\( \displaystyle 8k + 6 = -60,\quad k = -\frac{33}{4} \)
Since \(k \gt 0\), \(k = \frac{27}{4}\).