Question

A person standing at the junction of two straight paths represented by the equations $2x-3y+4=0$ and $3x+4y-5=0$. If he wants to reach the path whose equation is $6x-7y+8=0$ in the least time, then the equation of the path he should follow is

• A. $119x+102y=125$
• B. $129x+102y=125$
• C. $102x+119y=125$
• D. $129x+119y=124$

Answer 1

The correct option is A $119x+102y=125$

The equation of the given lines are
$2x-3y+4=0\cdots \left(i\right)$
$3x+4y-5=0\cdots \left(ii\right)$
$6x-7y+8=0\cdots \left(iii\right)$
Junction of $\left(i\right)$ and $\left(ii\right)$, solving both the equations, we get
$x=-\frac{1}{17},y=\frac{22}{17}$

Thus, the person is standing at
$(-\frac{1}{17},\frac{22}{17})$

The person can reach the path $\left(iii\right)$ in the least time if he walks along the perpendicular line to $\left(iii\right)$ from the junction point.
Now, line perpendicular to $\left(iii\right)$ is
$7x+6y+\lambda =0$
Putting $(-\frac{1}{17},\frac{22}{17})$, we get
$\frac{-7+132+17\lambda }{17}=0\Rightarrow \lambda =-\frac{125}{17}$
Therefore, the equation of line is
$7x+6y-\frac{125}{17}=0\Rightarrow 119x+102y=125$