Find values of k if area of triangle is 4 square units and vertices are

Solution:

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and

(x3, y3) is the absolute value of the determinant (Δ), where

$\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|$

It is given that the area of triangle is 4 square units.

$\therefore \Delta=\pm 4$

(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,

$\Delta=\frac{1}{2}\left|\begin{array}{lll}k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1\end{array}\right|$

$=\frac{1}{2}[k(0-2)-0(4-0)+1(8-0)]$

 

$=\frac{1}{2}[-2 k+8]=-k+4$

$\therefore-K+4=\pm 4$

When $-k+4=-4, k=8$.

When $-k+4=4, k=0$

 

Hence, $k=0,8$.

(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}-2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1\end{array}\right|$

$=\frac{1}{2}[-2(4-k)]$

$=k-4$

 

$\therefore k-4=\pm 4$

When $k-4=-4, k=0$.

When $k-4=4, k=8$.

 

Hence, $k=0,8$.