Let a triangle be bounded by the lines L_{1}: | vedclass

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(B) Point $A$ lies on $L_{2}: -4x + 3y = 12$. Let $A = (\alpha, \frac{12+4\alpha}{3})$.Point $B$ lies on $L_{1}: 2x + 5y = 10$. Let $B = (\beta, \frac

Vedclass · Accepted Answer

$\frac{132}{13}$

Vedclass · Answer

(B) Point $A$ lies on $L_{2}: -4x + 3y = 12$. Let $A = (\alpha, \frac{12+4\alpha}{3})$.Point $B$ lies on $L_{1}: 2x + 5y = 10$. Let $B = (\beta, \frac{10-2\beta}{5})$.Point $P(2, 3)$ divides $AB$ internally in the ratio $1:3$. Using the section formula:$2 = \frac{1 \cdot \beta + 3 \cdot \alpha}{1+3} \Rightarrow 3\alpha + \beta = 8$$3 = \frac{1 \cdot (\frac{10-2\beta}{5}) + 3 \cdot (\frac{12+4\alpha}{3})}{1+3}$ $\Rightarrow 12 = \frac{10-2\beta}{5} + 12 + 4\alpha$ $\Rightarrow 4\alpha - \frac{2\beta}{5} = -2$ $\Rightarrow 20\alpha - 2\beta = -10$ $\Rightarrow 10\alpha - \beta = -5$.Solving $3\alpha + \beta = 8$ and $10\alpha - \beta = -5$,we get $13\alpha = 3 \Rightarrow \alpha = \frac{3}{13}$ and $\beta = 8 - 3(\frac{3}{13}) = \frac{95}{13}$.Thus,$A = (\frac{3}{13}, \frac{56}{13})$ and $B = (\frac{95}{13}, -\frac{12}{13})$.Vertex $C$ is the intersection of $L_{1}$ and $L_{2}$. Solving $2x+5y=10$ and $-4x+3y=12$,we get $C = (-\frac{15}{13}, \frac{32}{13})$.Area of $	riangle ABC = \frac{1}{2} |x_{A}(y_{B}-y_{C}) + x_{B}(y_{C}-y_{A}) + x_{C}(y_{A}-y_{B})|$Area $= \frac{1}{2} |\frac{3}{13}(-\frac{12}{13} - \frac{32}{13}) + \frac{95}{13}(\frac{32}{13} - \frac{56}{13}) - \frac{15}{13}(\frac{56}{13} + \frac{12}{13})|$Area $= \frac{1}{2 \cdot 169} |3(-44) + 95(-24) - 15(68)| = \frac{1}{338} |-132 - 2280 - 1020| = \frac{3432}{338} = \frac{132}{13}$ sq. units.

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