#### Please solve RD Sharma class 12 chapter Determinants exercise 5.3 question 1 subquestion (iv) maths textbook solution

$Answer\! : 9sq.units.$

Hints: By putting the values of the co-ordinate in the formulaof area, we will calculate the area of the triangle.

$Given\, \! : \left ( 0,0 \right )\! ,\left ( 6,0 \right )\: and \: \left ( 4,3 \right )\, \! .$

$Explanation\! : V\! ertices \; are \left ( 0,0 \right )\! ,\left (6,0 \right )\! , \left ( 4,3 \right )\, \!.$

$Area\; o\! f\; triangles\; is =\! \Delta\! =\frac{1}{2}\begin{vmatrix} X_{1} &Y_{1} & 1\\ X_{2} &Y_{2} & 1\\ X_{3} & Y_{3}& 1 \end{vmatrix}$

$\Rightarrow where, \begin{matrix} X_{1}=0 &Y_{1}=0 & \\ X_{2}=6 &Y_{2}=0 & \\ X_{3}=4 &Y_{3}=3 & \end{matrix}$

$\Rightarrow \! \Delta \, \! =\frac{1}{2}\begin{vmatrix} 0 &0 &1 \\ 6 &0 &1 \\ 4 &3 &1 \end{vmatrix}$

$\Rightarrow \! \Delta \, \! =\frac{1}{2}\left ( 0\! \begin{vmatrix} 0 &1 \\ 3 &1 \end{vmatrix}-0\! \begin{vmatrix} 6 &1 \\ 4 &1 \end{vmatrix}+1\! \begin{vmatrix} 6 &0 \\ 4 &3 \end{vmatrix} \right )$

$=\frac{1}{2}\left [ 0\! \left ( 0-3 \right ) -0\! \left ( 6-4 \right )+1\! \left ( 18-0 \right )\right ]$

$=\frac{1}{2}\left [ 0-0+18\right ]$

$=\frac{1}{2}\times 18$

$=9sq. units$