Question

The equation of the straight line passing through the point of intersection of the lines $x-y=1$ and $2x-3y+1=0$ and parallel to the line $3x+4y=14$ is

• A. $3x+4y+24=0$
• B. $3x+4y-24=0$
• C. $4x+3y+24=0$
• D. $4x+3y-24=0$

Answer 1

The correct option is B $3x+4y-24=0$

Equation of any line through the point of intersection of lines
$x-y-1=0$ and $2x-3y+1=0$ is
$(x-y-1)+\lambda (2x-3y+1)=0$
$\Rightarrow (1+2\lambda )x-(1+3\lambda )y+(\lambda -1)=0\cdots \left(1\right)$
Slope of line
$m_1=\frac{1+2\lambda }{1+3\lambda }$
Slope of line $3x+4y-14=0$ is
$m_2=-\frac{3}{4}$
As both the lines are parallel, we get
$\frac{1+2\lambda }{1+3\lambda }=-\frac{3}{4}\Rightarrow 4+8\lambda =-3-9\lambda \Rightarrow \lambda =-\frac{7}{17}$

Putting the value of $\lambda$ in equation $\left(1\right)$, we get
$\frac{3}{17}x+\frac{4}{17}y-\frac{24}{17}=0\therefore 3x+4y-24=0$


Alternate solution:
Line parallel to $3x+4y-14=0$ is
$3x+4y+\mu =0$
Finding the point of intersection of the other two lines,
$x-y-1=0$ and $2x-3y+1=0$
$x=y+1\Rightarrow 2(y+1)-3y+1=0\Rightarrow y=3\Rightarrow x=4$
Now,
$3\times 4+4\times 3+\mu =0\Rightarrow \mu =-24$
Hence, the required equation of line is
$3x+4y-24=0$