#### Explain solution RD Sharma class 12 chapter 5 Determinants exercise multiple choise question 36 maths

Correct option(b)

Hint:

Use determinant to solve the area of triangle.

Given:

Area of a triangle = 9sq.units

Vertices of triangle = (-3,0),(3,0) and (0,k)

We have to find the value of k.

Solution:

We have,

Area of a triangle = 9sq.units

$\frac{1}{2}\left|\begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{array}\right|=9$

Solving the determinant, we get

$\left | -k(-6) \right |=18$

$\left | 6k \right |=18$

$k=\pm 3$

Hence k = 3 is required value of k.