Given, a person is standing at the junction of the lines,
2x−3y+4=0 ...(i)
3x+4y−5=0 ...(ii)
Solution of Eqs. (i) and (ii) will be the junction point, where the man is standing.
On solving Eqs. (i) and (ii),
x= and
y= ∴ Junction point
A=(,) Given, equation of path,
6x−7y+8=0 ...(iii)
The person can reach this path in the least time, if he walks along the perpendicular line to Eq. (iii)
from(,) ∴ Slope of line (iii)
=| −Coefficientofx |
| Coefficientofy |
== Hence, slope of the line perpendicular to line (iii)
m== ∴ Equation of required line passing through
(,) and having slope
is,
y−=(x+) 6(17y−22)=−7(17x+1) 102y−132=−119x−7 ⇒
119x+102y=125