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

The point of intersection of lines $2x-3y+4=0$ and $3x+4y-5=0$ is given by (-1,-2).

Since the shortest path through point A is perpendicular line AB,

$\therefore$ Slope of line $6x-7y+8=0is\frac{6}{7}$

So the slope of required line is $\frac{-7}{6}$

Thus equation of required line is

$y+2=\frac{-7}{6}(x+1)$

$\Rightarrow 6y+12=-7x-7$

$\Rightarrow 7x+6y+19=0$