Question

Find k, if the area of the triangle whose vertices are (2, 2), (6, 6) and (5, k) is 4

Answer 1

Refer the attachment.

Answer 2

Answer:

k=7

Step-by-step explanation:

   Given:

            $\left ( x_{1},y_{1} \right )$ = (2,2)

          $\left ( x_{2},y_{2} \right )$ = (6,6)

          $\left ( x_{3},y_{3} \right )$ = (5,k)

       area of triangle = 4

now,

    Area of triangle with given vertices = $\frac{1}{2}\left ( x_{1\left ( y_{2}-y_{3} \right )}+x_{2\left ( y_{3}-y_{1} \right )}+x_{3\left ( y_{1}-y_{2} \right )} \right )$

⇒ $4 =\frac{1}{2}\left ( 2\left ( 6-k\right )+6\left ( k-2\right )+ 5\left (2-6\right )\right )$

⇒ 8 = 4k-20

⇒ 4k = 28

⇒ k = 7