Find the value of $a$ for which the area of the triangle formed by the points $A (a, 2 a), B (- 2, 6)$ and $C (3, 1)$ is $10$ square units.

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Solution

Area of triangle having 3 vertices
$A = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣$

Here, $x_{1} = a, y_{1} = 2 a, x_{2} = - 2, y_{2} = 6, x_{3} = 3, y_{3} = 1$

$\Rightarrow$ $10 = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} a & 2 a & 1 - 2 & 6 & 1 3 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣$

$20 = a (6 - 1) - 2 a (- 2 - 3) + 1 (- 2 - 18)$

$20 = 5 a + 10 a - 20$

$15 a = 40$

$a = \frac{8}{3}$

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