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If the straight line $L \equiv 3 x+4 y-k=0$ cuts the line segment joining the points $P(2,-1)$ and $Q(1,1)$ in the ratio $4: 1$, then the equation of the line parallel to the line $y=x$ and concurrent with the lines $P Q$ and $L=0$ is
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Verified Answer
The correct answer is:
$5 x-5 y-3=0$
Point of intersection of given line $L=0$ and line segment $P Q$ is
$$
R\left(\frac{(4 \times 1)+(1 \times 2)}{5}, \frac{(4 \times(1))+(1 \times(-1))}{5}\right)=R\left(\frac{6}{5}, \frac{3}{5}\right)
$$
Now, point $R$ on the Line $L=0$,
So,
$$
\frac{18}{5}+\frac{12}{5}-k=0 \Rightarrow k=6
$$
Now, equation of Line concurrent with the lines $L=0$ and $P Q$ is
$\therefore$ Line Eq. (i) is parallel to the Line $y=x$, so
$$
\begin{aligned}
-\frac{3+2 \lambda}{4+\lambda} & =1 \\
\Rightarrow \quad \lambda & =-\frac{7}{3} .
\end{aligned}
$$
$\therefore$ Equation of required line is
$$
5 x-5 y-3=0
$$
$$
R\left(\frac{(4 \times 1)+(1 \times 2)}{5}, \frac{(4 \times(1))+(1 \times(-1))}{5}\right)=R\left(\frac{6}{5}, \frac{3}{5}\right)
$$
Now, point $R$ on the Line $L=0$,
So,
$$
\frac{18}{5}+\frac{12}{5}-k=0 \Rightarrow k=6
$$
Now, equation of Line concurrent with the lines $L=0$ and $P Q$ is
$\therefore$ Line Eq. (i) is parallel to the Line $y=x$, so
$$
\begin{aligned}
-\frac{3+2 \lambda}{4+\lambda} & =1 \\
\Rightarrow \quad \lambda & =-\frac{7}{3} .
\end{aligned}
$$
$\therefore$ Equation of required line is
$$
5 x-5 y-3=0
$$
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