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

Solution:

The equations of the given lines are

2x – 3y + 4 = 0 … (1)

3x + 4y – 5 = 0 … (2)

6x – 7y + 8 = 0 … (3)

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

On solving equations $(1)$ and $(2)$, we obtain $x=-\frac{1}{17}$ and $y=\frac{22}{17}$.

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

The person can reach path $(3)$ in the least time if he walks along the perpendicular line to $(3)$ from point $\left(-\frac{1}{17}, \frac{22}{17}\right)$

Slope of the line $(3)=\frac{6}{7}$

$\therefore$ Slope of the line perpendicular to line $(3)=-\frac{1}{\left(\frac{6}{7}\right)}=-\frac{7}{6}$

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

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

$6(17 y-22)=-7(17 x+1)$

$102 y-132=-119 x-7$

$119 x+102 y=125$

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