A person standing at the junction (crossing) of two straight paths represented by the equations

It is given that

2x – 3y + 4 = 0 …… (1)

3x + 4y – 5 = 0 ……. (2)

6x – 7y + 8 = 0 …… (3)

Here the person is standing at the junction of the paths represented by lines (1) and (2).

By solving equations (1) and (2) we get

x = \(- \dfrac{1}{17}\) and y = \( \dfrac{22}{17}\)

Hence, the person is standing at point (\( - \dfrac{1}{17}\),\( \dfrac{22}{17}\)).

We know that the person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point 

(\( - \dfrac{1}{17}\),\( \dfrac{22}{17}\))

Here the slope of the line (3) = \( \dfrac{6}{7}\)

We get the slope of the line perpendicular to line (3)

 = \( -\dfrac{1}{\dfrac{6}{7}}= - \dfrac{7}{6}\)

So the equation of line passing through 

(\( - \dfrac{1}{17}\),\( \dfrac{22}{17}\)) and having a slope of \( - \dfrac{7}{6}\) is written as

\(( y- \dfrac{22}{17})= - \dfrac{7}{6}(x+ \dfrac{1}{17})\)

By further calculation

6 (17y – 22) = – 7 (17x + 1)

By multiplication

102y – 132 = – 119x – 7

We get

1119x + 102y = 125

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