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

$Answer\! : \frac{47}{2}sq.units$

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

$Given\, \! : \left ( 2,7 \right )\! ,\left ( 1,1 \right )\: and \: \left ( 10,8 \right )\, \! .$

$Explanation\! : V\! ertices\: are \left ( 2,7 \right )\! ,\left ( 1,1 \right )\! , \left ( 10,8 \right )\, \! .$

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

$\Rightarrow where, \begin{matrix} X_{1}=2 &Y_{1}=7 & \\ X_{2}=1 &Y_{2}=1 & \\ X_{3}=10 &Y_{3}=8 & \end{matrix}$

$\Rightarrow \! \Delta \, \! =\frac{1}{2}\begin{vmatrix} 2 &7 &1 \\ 1 &1 &1 \\ 10 &8 &1 \end{vmatrix}$

$\Rightarrow \! \Delta \, \! =\frac{1}{2}\left ( 2\! \begin{vmatrix} 1 &1 \\ 8 &1 \end{vmatrix}-7\! \begin{vmatrix} 1 &1 \\ 10 &1 \end{vmatrix}+1\! \begin{vmatrix} 1 &1 \\ 10 &8 \end{vmatrix} \right )$

$=\frac{1}{2}\left [ 2\! \left ( 1-8 \right ) -7\! \left ( 1-10 \right )+1\! \left ( 8-10 \right )\right ]$

$=\frac{1}{2}\left [ 2\! \left ( -7 \right )-7\! \left ( -9 \right )+1\! \left ( -2 \right ) \right ]$

$=\frac{1}{2}\left [ -14+63-2 \right ]$

$=\frac{47}{2}sq. units$

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