#### Please solve RD Sharma class 12 chapter Determinants exercise 5.3 question 6 maths textbook solution

Answer: $x$ = -2

Hints: Using determinant, find area of triangle and find value of $x$.

$\boldsymbol{Given\: \! :} \left ( x,4 \right )\! , \left ( 2,-6 \right )\; and\; \left ( 5,4 \right ).$

area of triangle = 35 sq. Units.

$\boldsymbol{Explanation:} V\! ertices\; are \left ( x,4 \right )\! , \left ( 2,-6 \right )\; and\; \left ( 5,4 \right ).$

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

$\Rightarrow \Delta=\frac{1}{2}\left|\begin{array}{lll} x & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1 \end{array}\right|=35$

$\Delta =\frac{1}{2}\left ( x\! \begin{vmatrix} -6 & 1\\ 4 & 1 \end{vmatrix}-4\! \begin{vmatrix} 2 & 1\\ 5 & 1 \end{vmatrix}+1\! \begin{vmatrix} 2 & -6\\ 5 & 4 \end{vmatrix} \right )=35$

$\Rightarrow \frac{1}{2}[x(-6-4)-4(2-5)+1(8-(-30))]=35$

$\Rightarrow \frac{1}{2}[x(-10)-4(-3)+1(38)]=35$

$\Rightarrow \frac{1}{2}[-10x+12+38]=35$

$\Rightarrow -10x+50=70$

$\Rightarrow -10x=70-50$

$\Rightarrow x=-2$