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  • Let a triangle be bounded by the lines <img alt="\mathrm{L}_{1}: 2 x+5 y=10 ; \mathrm{L}_{2}:-4 x+3 y=12" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BL%7D_%7B1%7D%3A%202%20x&am

Let a triangle be bounded by the lines \mathrm{L}_{1}: 2 x+5 y=10 ; \mathrm{L}_{2}:-4 x+3 y=12 and the line \mathrm{L}_{3}, which passes through the point \mathrm{P}(2,3), intersects \mathrm{L}_{2} at \mathrm{A}$ and $\mathrm{L}_{1}$ at $\mathrm{B} . If the point \mathrm{P} divides the line-segment \mathrm{A B}, internally in the ratio \mathrm{ 1: 3, }then the area of the triangle is equal to:

Option: 1

\frac{110}{13}


Option: 2

\frac{132}{13}


Option: 3

\frac{142}{13}


Option: 4

\frac{151}{13}


Answers (1)

best_answer

\mathrm{Let \: A \: be\: \left(a, 4+\frac{4 a}{3}\right)}        \mathrm{\left(\because A \text { lies on } L_{2}\right)}

\mathrm{and \: B\: be \left(b, 2-\frac{2}{5} b\right)}        \mathrm{(\because B\: lies\: on\: L_1)}

Using section formula for point P

\mathrm{2=\frac{3 a+b}{4} \text { and } 3=\frac{3\left(4+\frac{4 a}{3}\right)+\left(2-\frac{2 b}{5}\right)}{4}}

\mathrm{\Rightarrow 3 a+b=8 \text { and } 4 a-\frac{2 b}{5}=-2} \\

\mathrm{\Rightarrow \quad a=\frac{3}{13}, b=\frac{95}{13}} \\

\mathrm{\therefore \quad A \text { is }\left(\frac{3}{13}, \frac{56}{13}\right), B \text { is }\left(\frac{95}{13}, \frac{-12}{13}\right) }

\mathrm{\therefore \text { Area of } \triangle A B C=\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| }                                     \begin{aligned} &=\frac{1}{2} \cdot \frac{1}{13^{3}}\left|\begin{array}{rrr} 3 & 56 & 1 \\ 95 & -12 & 1 \\ -15 & 32 & 1 \end{array}\right| \\ & \end{aligned}

=\frac{132}{13}

Hence the correct answer is option 2

Posted by

himanshu.meshram

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