If the area of a triangle whose vertices are $(-1, 3), (1, -5)$ and $(k,2)$ where $k > 0$ is 30 sq. units, then the value of $k$ is

$\frac{27}{4}$

The correct answer is Option (4) → $\frac{27}{4}$

Vertices: $A(-1,3),\ B(1,-5),\ C(k,2)$

Area of triangle formula:

$\text{Area} = \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$30 = \frac{1}{2}\left|(-1)(-5-2) + 1(2-3) + k(3 - (-5))\right|$

$30 = \frac{1}{2}\left|(7 - 1 + 8k)\right|$

$30 = \frac{1}{2}\left|8k + 6\right|$

$60 = |8k + 6|$

Case 1: $8k + 6 = 60 \Rightarrow k = \frac{54}{8} = \frac{27}{4}$

Case 2: $8k + 6 = -60 \Rightarrow k = -\frac{66}{8} = -\frac{33}{4}$

Given $k > 0$

$k = \frac{27}{4}$