Question

The equation of the straight line passing through the point of intersection of $x+2y=5$ and $3x+7y=17$ and perpendicular to the straight line $3x+4y=10$ is

• A. $4x-3y+1=0$
• B. $4x-3y+12=0$
• C. $4x-3y-2=0$
• D. $4x-3y+2=0$

Answer 1

The correct option is D $4x-3y+2=0$

Equation of line passing through the intersection of the lines $x+2y-5=0$ and $3x+7y-17=0$ is

$(x+2y-5)+\lambda (3x+7y-17)=0$
$\Rightarrow x(3\lambda +1)+y(7\lambda +2)-(17\lambda +5)=0\cdots \left(1\right)$
Slope of the line
$m_1=-\left(\frac{3\lambda +1}{7\lambda +2}\right)$
Now, slope of the other given line
$3x+4y-10=0m_2=-\frac{3}{4}$

As the lines are perpendicular, so
$m_1{m}_2=-1\Rightarrow -\left(\frac{3\lambda +1}{7\lambda +2}\right)(-\frac{3}{4})=-1\Rightarrow 9\lambda +3=-28\lambda -8\Rightarrow \lambda =-\frac{11}{37}$

Putting the value of $\lambda$ in equation $\left(1\right)$, we get
$x(-\frac{33}{37}+1)+y(-\frac{77}{37}+2)-(-\frac{187}{37}+5)=0$
$\therefore 4x-3y+2=0$


Alternate solution:
Given line $3x+4y-10=0$
Perpendicular line to this
$4x-3y+\mu =0\cdots \left(2\right)$
Now finding the point of intersection of
$x+2y-5=0$ and $3x+7y-17=0$
We get
$x=5-2y\Rightarrow 3(5-2y)+7y-17=0\Rightarrow y=2\Rightarrow x=1$
Using equation $\left(2\right)$, we get
$4-6+\mu =0\Rightarrow \mu =2$
So, the equation of the required line is
$4x-3y+2=0$