If the area of a triangle whose vertices are (-2, 4), (2, -6) and (k, 4), (k > 0) is 35 sq.units, then the value of k is

5

The correct answer is Option (2) → 5

$\text{Vertices }A(-2,4),\;B(2,-6),\;C(k,4),\;k>0$

$\text{Since }A\text{ and }C\text{ have }y=4,\;\overline{AC}\text{ is horizontal with length }|k-(-2)|=k+2.$

$\text{Perpendicular height from }B(2,-6)\text{ to the line }y=4\text{ is }|{-6}-4|=10.$

$\text{Area}=\frac{1}{2}\times(k+2)\times10=5(k+2).$

$5(k+2)=35\;\Rightarrow\;k+2=7\;\Rightarrow\;k=5.$

Final Answer: $k={5}$