JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 12)
Let a triangle be bounded by the lines L1 : 2x + 5y = 10; L2 : $$-$$4x + 3y = 12 and the line L3, which passes through the point P(2, 3), intersects L2 at A and L1 at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :
$${{110} \over {13}}$$
$${{132} \over {13}}$$
$${{142} \over {13}}$$
$${{151} \over {13}}$$
Explanation
$L_{1}: 2 x+5 y=10$
$L_{2}: -4 x+ 3 y=12$
_28th_June_Evening_Shift_en_12_1.png)
Solving $L_{1}$ and $L_{2}$ we get
$$ C \equiv\left(\frac{-15}{13}, \frac{32}{13}\right) $$
Now, Let $A\left(x_{1}, \frac{1}{3}\left(12+4 x_{1}\right)\right)$ and
$$ \begin{aligned} &B\left(x_{2}, \frac{1}{5}\left(10-2 x_{2}\right)\right) \\\\ &\therefore \quad \frac{3 x_{1}+x_{2}}{4}=2 \\\\ &\text { and } \frac{\left(12+4 x_{1}\right)+\frac{10-2 x_{2}}{5}}{4}=3 \end{aligned} $$
So, $3 x_{1}+x_{2}=8$ and $10 x_{1}-x_{2}=-5$
So, $\left(x_{1}, x_{2}\right)=\left(\frac{3}{13}, \frac{95}{13}\right)$
$$ \begin{aligned} &A=\left(\frac{3}{13}, \frac{56}{13}\right) \text { and } B=\left(\frac{95}{13}, \frac{-12}{13}\right) \\\\ &=\left|\frac{1}{2}\left(\frac{3}{13}\left(\frac{-44}{13}\right) \frac{-56}{13}\left(\frac{110}{13}\right)+1\left(\frac{2860}{169}\right)\right)\right| \\\\ &=\frac{132}{13} \text { sq. units } \end{aligned} $$
$L_{2}: -4 x+ 3 y=12$
Solving $L_{1}$ and $L_{2}$ we get
$$ C \equiv\left(\frac{-15}{13}, \frac{32}{13}\right) $$
Now, Let $A\left(x_{1}, \frac{1}{3}\left(12+4 x_{1}\right)\right)$ and
$$ \begin{aligned} &B\left(x_{2}, \frac{1}{5}\left(10-2 x_{2}\right)\right) \\\\ &\therefore \quad \frac{3 x_{1}+x_{2}}{4}=2 \\\\ &\text { and } \frac{\left(12+4 x_{1}\right)+\frac{10-2 x_{2}}{5}}{4}=3 \end{aligned} $$
So, $3 x_{1}+x_{2}=8$ and $10 x_{1}-x_{2}=-5$
So, $\left(x_{1}, x_{2}\right)=\left(\frac{3}{13}, \frac{95}{13}\right)$
$$ \begin{aligned} &A=\left(\frac{3}{13}, \frac{56}{13}\right) \text { and } B=\left(\frac{95}{13}, \frac{-12}{13}\right) \\\\ &=\left|\frac{1}{2}\left(\frac{3}{13}\left(\frac{-44}{13}\right) \frac{-56}{13}\left(\frac{110}{13}\right)+1\left(\frac{2860}{169}\right)\right)\right| \\\\ &=\frac{132}{13} \text { sq. units } \end{aligned} $$
Comments (0)
