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Let a triangle be bounded by the lines $L _{1}: 2 x +5 y =10$; $L _{2}:-4 x +3 y =12$ and the line $L _{3}$, which passes through the point $P (2,3)$, intersect $L _{2}$ at $A$ and $L _{1}$ at $B$. If the point $P$ divides the line-segment $A B$, internally in the ratio $1: 3$, then the area of the triangle is equal to
$\frac{110}{13}$
$\frac{132}{13}$
$\frac{142}{13}$
$\frac{151}{13}$
Solution
Points $A$ lies on $L _{2}$
$A \left(\alpha, 4+\frac{4}{3} \alpha\right)$
Points $B$ lies on $L _{1}$
$B \left(\beta, 2-\frac{2}{5} \beta\right)$
Points $P$ divides $AB$ internally in the ratio $1: 3$
$\Rightarrow P(2,3)=P\left(\frac{3 \alpha+\beta}{4}, \frac{3\left(4+\frac{4}{3} \alpha\right)+1\left(2-\frac{2}{5} \beta\right)}{4}\right)$
$\Rightarrow \alpha=\frac{3}{13}, \beta=\frac{95}{13}$
Point $A \left(\frac{3}{13}, \frac{56}{13}\right), B \left(\frac{95}{13},-\frac{12}{13}\right)$
Vertex $C$ of triangle is the point of intersection of $L _{1} and L _{2}$
$\Rightarrow C \left(-\frac{15}{13}, \frac{32}{13}\right)$
area $\triangle ABC =\frac{1}{2}\left\|\begin{array}{ccc}\frac{3}{13} & \frac{56}{13} & 1 \\ \frac{95}{13} & -\frac{12}{13} & 1 \\ -\frac{15}{13} & \frac{32}{13} & 1\end{array}\right\|$
$\frac{1}{2 \times 13^{3}}\left\|\begin{array}{ccc} 3 & 56 & 13 \\ 95 & -12 & 13 \\ -15 & 32 & 13 \end{array}\right\|$
area $\triangle ABC =\frac{132}{13}$ sq. units.
