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

$Answer\! : \frac{75}{2}sq.units$

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

$Given\, \! : \left ( 3,8 \right )\! ,\left ( -4,2 \right )\: and \: \left ( 5,-1 \right )\, \! .$

$Explanation\! : V\! ertices \; are \left ( 3,8 \right )\! ,\left ( -4,2 \right )\! , \left ( 5,-1 \right )\, \! .$

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

$\Rightarrow where, \begin{matrix} X_{1}=3 &Y_{1}=8 & \\ X_{2}=-4 &Y_{2}=2 & \\ X_{3}=5 &Y_{3}=-1 & \end{matrix}$

$\Rightarrow \! \Delta \, \! =\frac{1}{2}\begin{vmatrix} 3 &8 &1 \\ -4 &2 &1 \\ 5 &-1 &1 \end{vmatrix}$

$\Rightarrow \! \Delta \, \! =\frac{1}{2}\left ( 3\! \begin{vmatrix} 2 &1 \\ -1 &1 \end{vmatrix}-8\! \begin{vmatrix} -4 &1 \\ 5 &1 \end{vmatrix}+1\! \begin{vmatrix} -4 &2 \\ 5 &-1 \end{vmatrix} \right )$

$=\frac{1}{2}\left [ 3\! \left ( 2-\left ( -1 \right ) \right ) -8\! \left ( -4-\left ( 5 \right ) \right )+1\! \left ( 4-10 \right )\right ]$

$=\frac{1}{2}\left [ 3\! \left ( 2+1 \right )-8\! \left ( -9 \right )+1\! \left ( -6 \right ) \right ]$

$=\frac{1}{2}\left [ 3\! \left ( 3 \right )-8\! \left ( -9 \right )+1\! \left ( -6 \right ) \right ]$

$=\frac{1}{2}\left [ 9+72-6 \right ]$

$=\frac{75}{2}sq. units$