Question

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

Answer 1

Step 1- Finding the coordinates of the point where a person is standing

A person is standing at the junction of the below lines

$2x-3y+4=0eq\left(1\right)$

$3x+4y-5=0eq\left(2\right)$

Solving $eq\left(1\right)$ and $eq\left(2\right)$

$x=\frac{-1}{17}$ and $y=\frac{22}{17}$

the person is standing at a point $\left(\frac{-1}{17},\frac{22}{17}\right)$

Step 2 - Slope of the line perpendicular to the given path

The equation of the path is:

$6x-7y+8=0eq\left(3\right)$

When the person walks perpendicular to $eq\left(3\right)$ from point $\left(\frac{-1}{17},\frac{22}{17}\right)$ the person will reach in the least time

$$\begin{gathered} \because m_1\times m_2=-1 \\ \Rightarrow \frac{6}{7}\times m_2=-1 \\ \Rightarrow m_2=\frac{-7}{6} \end{gathered}$$

the slope of the line perpendicular to the $eq\left(3\right)=\frac{-7}{6}$

Step 3 - Equation of path that person should follow

The equation of the line passing through $\left(\frac{-1}{17},\frac{22}{17}\right)$and with the slope of $\left(\frac{-7}{6}\right)$is given as

$\begin{array}{rcl}& \Rightarrow & \left(y-\frac{22}{17}\right)=\frac{-7}{6}\left(x+\frac{1}{17}\right)\\ & \Rightarrow & 6\left(\frac{17y-22}{17}\right)=-7\left(\frac{17x+1}{17}\right)\\ & \Rightarrow & 102y-132=-119x-7\\ & \Rightarrow & 119x+102y=132-7\\ & \Rightarrow & 119x+102y=125\end{array}$

Therefore, the path that the person should follow is $119x+102y=125$