Let a triangle be bounded by the lines L _{1} | vedclass

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Question

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\rig

Vedclass · Accepted Answer

$\frac{132}{13}$

Vedclass · Answer

Points $A$ lies on $L _{2}$

$A \left(\alpha, 4+\frac{4}{3} \alphaight)$

Points $B$ lies on $L _{1}$

$B \left(\beta, 2-\frac{2}{5} \betaight)$

Points $P$ divides $AB$ internally in the ratio $1: 3$

$\Rightarrow P(2,3)=P\left(\frac{3 \alpha+\beta}{4}, \frac{3\left(4+\frac{4}{3} \alphaight)+1\left(2-\frac{2}{5} \betaight)}{4}ight)$

$\Rightarrow \alpha=\frac{3}{13}, \beta=\frac{95}{13}$

Point $A \left(\frac{3}{13}, \frac{56}{13}ight), B \left(\frac{95}{13},-\frac{12}{13}ight)$

Vertex $C$ of triangle is the point of intersection of $L _{1} and L _{2}$

$\Rightarrow C \left(-\frac{15}{13}, \frac{32}{13}ight)$

area $	riangle 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}ight\|$

$\frac{1}{2 	imes 13^{3}}\left\|\begin{array}{ccc} 3 & 56 & 13 \ 95 & -12 & 13 \ -15 & 32 & 13 \end{array}ight\|$

area $	riangle ABC =\frac{132}{13}$ sq. units.

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