#### Provide solution for RD Sharma maths class 12 chapter Determinants exercise 5.3 question 13 subquestion (ii)

Answer: $k$=8

Hints: Putting the values of vertices, in the formula of area of triangle.

Given: (-2,0), (0,4) and (0,$k$)
area of triangle= 4 sq. units

Explanation: Vertices are (-2,0), (0,4) and (0,$k$)

$\text { Area of triangle is }=\Delta=\frac{1}{2}\left|\begin{array}{lll} X_{1} & Y_{1} & 1 \\ X_{2} & Y_{2} & 1 \\ X_{3} & Y_{3} & 1 \end{array}\right|=4$

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

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

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

$\Rightarrow -8+2k=8$

$\Rightarrow 2k=16$

$\Rightarrow k=8$