If the area of a triangle formed by the points (k, 2k) (-2, 6) and (3, 1) is 20 square units. Find the value of k.

Area of a triangle with vertices $({ x }_{ 1 },{ y }_{ 1 })$ ,$({ x }_{ 2 },{ y }_{ 2 })$ and $({ x }_{ 3 },{ y }_{ 3 })$ is $\left| \frac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

Hence, area of the triangle with given vertices $\left| \frac { k(6-1)-2(1-2k)+3(2k-6) }{ 2 } \right| = 20$

$\Longrightarrow \left| \frac { 6k-k-2+4k+6k-18 }{ 2 } \right| = 20$

$\left| \frac { 15k-20 }{ 2 } \right| = 20$

$\left| 15k-20 \right| = 40$

$15k - 20 = 40$

$15k = 60$

$k = 4$