Question

A person standing at the junction (crossing) 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 equation of the path that he should follow.

Answer 1

Simplification of given data
Let equation of lines
$OA:2x-3y+4=0\dots \left(i\right)$
$OB:3x+4y-5=0\dots \left(ii\right)$
$AB:6x-7y+8=0\dots \left(iii\right)$
The two paths cross at point $O$
$\therefore$ The person is standing at point $O$
From point $O$, if he has to reach line $AB$ in least time, Least distance from point $O$ to line $AB$ will be perpendicular distance.

Solve for coordinate of point $O$
Point $O$ is the intersection of $OA\&OB$
Calculating, $\left(i\right)\times 4+\left(ii\right)\times 3$
$\Rightarrow (8x-12y+16)+(9x+12y-15)=0$
$\Rightarrow 17x+1=0$
$\Rightarrow x=-\frac{1}{17}$
Now, substituting $x=-\frac{1}{17}$ in $\left(i\right)$ we get
$-\frac{2}{17}-3y+4=0$
$\Rightarrow 3y=\frac{-2+68}{17}$
$\Rightarrow 3y=\frac{66}{17}$
$\Rightarrow y=\frac{22}{17}$
Thus, the coordinate of point O is $(\frac{-1}{17},\frac{22}{17})$

Equation of line $OM$
Now, line $OM$ is perpendicular to line $AB$
Equation of $AB$ is
$\Rightarrow 6x-7y+8=0$
$\Rightarrow 6x+8=7y$
$\Rightarrow 7y=6x+8$
$\Rightarrow y=\frac{6x+8}{7}$
$\Rightarrow y=\left(\frac{6}{7}\right)x+\frac{8}{7}$

The above equation is of the form $y=mx+c$
Slope of $AB=\frac{6}{7}$
When two lines are perpendicular then product of their slopes is $-1$.
Thus, Slope of $OM\times \text{Slope of}AB=-1$
Slope of $OM=\frac{-1}{\text{Slope of AB}}=\frac{-1}{\frac{6}{7}}=\frac{-7}{6}$
Equation of line $OM$ passing through $O(\frac{-1}{17},\frac{22}{17})$ having slope $\frac{-7}{6}$ is

$y-\frac{22}{17}=\frac{-7}{6}(x-\left(\frac{-1}{17}\right))$

$\Rightarrow y-\frac{22}{17}=\frac{-7}{6}(x+\frac{1}{17})$

$\Rightarrow y-\frac{22}{17}=\frac{-7}{6}x-\frac{7}{6}\times \frac{1}{17}$

$\Rightarrow y-\frac{22}{17}=\frac{-7}{6}x-\frac{7}{6\times 17}$

$\Rightarrow y+\frac{7}{6}x=\frac{22}{17}-\frac{7}{6\times 17}$

$\Rightarrow \frac{6y+7x}{6}=\frac{22\left(6\right)-7}{6\times 17}$

$\Rightarrow \frac{6y+7x}{6}=\frac{132-7}{6\times 17}$

$\Rightarrow \frac{6y+7x}{6}=\frac{125}{6\times 17}$

$\Rightarrow 6y+7x=\frac{125}{17}$

$\Rightarrow 17(6y+7x)=125$

$\Rightarrow 17\left(6y\right)+17\left(7x\right)=125$

$\Rightarrow 102y+119x=125$

$\therefore$ The path he should follow is $119x+102y=125$.