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

$Answer\! : 15sq.units.$

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

$Given\, \! : \left ( -1,-8 \right )\! ,\left ( -2,-3 \right )\: and \: \left ( 3,2 \right )\, \! .$

$Explanation\! : V\! ertices \; are \left ( -1,-8 \right )\! ,\left ( -2,-3 \right )\! , \left ( 3,2 \right )\, \! .$

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

$\Rightarrow where, \begin{matrix} X_{1}=-1 &Y_{1}=-8 & \\ X_{2}=-2 &Y_{2}=-3 & \\ X_{3}=3 &Y_{3}=2 & \end{matrix}$

$\Rightarrow \! \Delta \, \! =\frac{1}{2}\begin{vmatrix} -1 &-8 &1 \\ -2 &-3 &1 \\ 3 &2 &1 \end{vmatrix}$

$\Rightarrow \! \Delta \, \! =\frac{1}{2}\left ( -1\! \begin{vmatrix} -3 &1 \\ 2 &1 \end{vmatrix}-\left ( -8 \right )\! \begin{vmatrix} -2 &1 \\ 3 &1 \end{vmatrix}+1\! \begin{vmatrix} -2 &-3 \\ 3 &2 \end{vmatrix} \right )$

$=\frac{1}{2}\left [ -1\! \left ( -3-2 \right ) +8\! \left ( -2-3 \right )+1\! \left ( -4+9 \right )\right ]$

$=\frac{1}{2}\left [ -1\! \left ( -5 \right )+8\! \left ( -5 \right )+1\! \left ( 5\right ) \right ]$

$=\frac{1}{2}\left [ 5-40+5\right ]$

$=\frac{1}{2}\times 30$

$=15sq. units$