#### Provide solution for RD Sharma maths class 12 chapter Determinants exercise 5.3 question 7

$\boldsymbol{Answer\! : }\frac{+13}{2} sq. units.\;\; N\! o, \; points\; are\; not\; collinear.$

Hints: First find the determinant and then area of the triangle. If result found be zero, then given points are collinear.

Given: Vertices are (1,4), (2,3) and (-5, -3).

Explanation: Vertices are (1,4), (2,3) and (-5, -3).

$\text { Area of triangle is }=\Delta=\frac{1}{2}\left|\begin{array}{lll} 1 & 4 & 1 \\ 2 & 3 & 1 \\ -5 & -3 & 1 \end{array}\right|$

$\Rightarrow|\Delta|=\frac{1}{2}\left(1\left|\begin{array}{cc} 3 & 1 \\ -3 & 1 \end{array}\right|-4\left|\begin{array}{cc} 2 & 1 \\ -5 & 1 \end{array}\right|+1\left|\begin{array}{cc} 2 & 3 \\ -5 & -3 \end{array}\right|\right)$

$\Rightarrow \frac{1}{2}[1(3-(-3))-4(2-(-5))+1(-6-(-15))]$

$\Rightarrow \frac{1}{2}[1(6)-4(7)+1(9)]$

$\Rightarrow \frac{1}{2}[6-28+9]$

$\Rightarrow \frac{1}{2}\times (-13)$

$\Rightarrow \left | \frac{-13}{2} \right | sq. units \Rightarrow \frac{13}{2} sq. units$

and determinant is not zero, that is why points are not collinear.