Let a triangle be bounded by the lines L_1: 2x+5y=10; L_2: -4x+3y=12 and the line L_3, whi…

Question

Let a triangle be bounded by the lines $L_1: 2x+5y=10$; $L_2: -4x+3y=12$ and the line $L_3$, which passes through the point $P(2,3)$, intersects $L_2$ at $A$ and $L_1$ 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:

Accepted Answer

$L_{1}: 2 x+5 y=10$

$L_{2}: -4 x+ 3 y=12$

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, $\left(x_{1}, x_{2}\right)=\left(\frac{3}{13}, \frac{95}{13}\right)$

$$
\begin{aligned}
&A=\left(\frac{3}{13}, \frac{56}{13}\right) \text { and } B=\left(\frac{95}{13}, \frac{-12}{13}\right) \\
&=\left|\frac{1}{2}\left(\frac{3}{13}\left(\frac{-44}{13}\right) \frac{-56}{13}\left(\frac{110}{13}\right)+1\left(\frac{2860}{169}\right)\right)\right| \\
&=\frac{132}{13} \text { sq. units }
\end{aligned}
$$

Solutions · 1