Q63 / 120 · VITEEE 2012 · Mathematics
The equation of straight line through the intersection of the lines $x-2 y=1$ and $x+3 y=2$ and parallel $3 x+4 y=0$ is
- A$3 x+4 y+5=0$
- B$3 x+4 y-10=0$
- Correct answerC$3 x+4 y-5=0$
- D$3 x+4 y+6=0$
Solution
The intersection point of lines $x-2 y=1$ and $x+3 y=2$ is $\left(\frac{7}{5}, \frac{1}{5}\right)$
Since, required line is parallel to $3 x+4 y=0$.
Therefore, the slope of required line is $-\frac{3}{4}$.
$\therefore$ Equation of required line which passes through $\left(\frac{7}{5}, \frac{1}{5}\right)$ is given by
$$
\begin{array}{l}
y-\frac{1}{5}=-\frac{3}{4}\left(x-\frac{7}{5}\right) \\
\Rightarrow \frac{3 x}{4}+y=\frac{21}{20}+\frac{1}{5} \\
\Rightarrow 3 x+4 y-5=0
\end{array}
$$
Hence, the correct answer is option: 'C'.