6. Find the value of x if the area of $\vartriangle $ is $35$ square cms with vertices $(x,4)$, $(2,-6)$ and $(5,4)$.

Home » 6. Find the value of x if the area of is square cms with vertices , and .

6. Find the value of x if the area of $\vartriangle$ is $35$ square cms with vertices $(x,4)$ , $(2,-6)$ and $(5,4)$ .

6. Find the value of x if the area of $\vartriangle$ is $35$ square cms with vertices $(x,4)$ , $(2,-6)$ and $(5,4)$ .

As per the question it is given that, $(x,4)$ , $(2,-6)$ and $(5,4)$ are the vertices of a triangle.

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\And \left( {{x}_{3}},{{y}_{3}} \right)$ , then the area of the triangle is given by,

$\vartriangle =\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|=0$

Then putting the value in above formula,

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

Removing Modulus

$\pm 2\times 35=\left| \begin{matrix} x & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1 \\ \end{matrix} \right|$

Expanding along ${{R}_{1}}$

$\Rightarrow \left[ x\left| \begin{matrix} -6 & 1 \\ 4 & 1 \\ \end{matrix} \right|-4\left| \begin{matrix} 2 & 1 \\ 5 & 1 \\ \end{matrix} \right|+1\left| \begin{matrix} 2 & -6 \\ 5 & 4 \\ \end{matrix} \right| \right]=\pm 70$