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

Given a person is standing at the junction of the below lines

$2x-3y+4=0$ ...(1)
$3x+4y-5=0$ ...(2)
Solving equation (1) and (2) ,we get

$x=-\frac{1}{17}$ and $y=\frac{22}{17}$
So, the person is standing at point $(-\frac{1}{17},\frac{22}{17})$

Given equation of path is

$6x-7y+8=0$ ...(3)
The person can reach this path in the least time if he walks along the perpendicular line to (3) from point $(-\frac{1}{17},\frac{22}{17})$

Slope of the line (3)$=\frac{6}{7}$
$\therefore$ slope of the line perpendicular to line (3)$=-\frac{1}{\frac{6}{7}}=-\frac{7}{6}$
The equation of the line passing through $(-\frac{1}{17},\frac{22}{17})$ and having a slope of $-\frac{7}{6}$ is given by
$y-\frac{22}{17}=-\frac{7}{6}(x+\frac{1}{17})$
$6(17y-22)=-7(17x+1)$
$102y-132=-119x-7$
$119x+102y=125$
Hence, the path that person should follow is $119x+102y=125$