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

The triangle has vertices at $(-1, 3)$, $(1, -5)$, and $(k, 2)$ where $k > 0$.

The area of the triangle is $30$ sq. units.

For vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

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

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Assign the vertices:

$(x_1, y_1) = (-1, 3)$

$(x_2, y_2) = (1, -5)$

$(x_3, y_3) = (k, 2)$

Substituting into the formula:

$\text{Area} = \frac{1}{2}|(-1)[(-5) - 2] + (1)[2 - 3] + k[3 - (-5)]|$

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Simplifying each term:

$(-1)[(-5) - 2] = (-1)(-7) = 7$

$(1)[2 - 3] = (1)(-1) = -1$

$k[3 - (-5)] = k[8] = 8k$

Combining:

$\text{Area} = \frac{1}{2}|7 - 1 + 8k|$

$\text{Area} = \frac{1}{2}|6 + 8k|$

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The area equals $30$ sq. units:

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

$|6 + 8k| = 60$

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The absolute value equation gives two cases:

Case 1: $6 + 8k = 60$

$8k = 54$

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

Case 2: $6 + 8k = -60$

$8k = -66$

$k = -\frac{33}{4}$

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Since $k > 0$, the value $k = \frac{27}{4}$ is valid while $k = -\frac{33}{4}$ is rejected.

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