A person standing at the junction (crossing) of 2 straight paths represented by the equations 2x − 3y + 4 =…
A person standing at the junction (crossing) of 2 straight paths represented by the equations 2x − 3y + 4 = 0 and 3x + 4 y − 5 = 0, wants to reach the path whose equation is 6 x − 7 y + 8 = 0 in the least time, then the equation of the path he should follow is
119x − 102 y − 125 = 0
119x + 102 y − 125 = 0
102 x + 119y − 125 = 0
102 x − 119y + 125 = 0
Solution
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), $x=\frac{-1}{17}$ and $y=\frac{22}{17}$
$\therefore$ Junction point $A=\left(\frac{-1}{17}, \frac{22}{17}\right)$
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)
$\begin{aligned} & \text { from }\left(\frac{-1}{17}, \frac{22}{17}\right) . \\ & \therefore \text { Slope of line (iii) }=\frac{- \text { Coefficient of } x}{\text { Coefficient of } y}=\frac{-6}{-7}=\frac{6}{7}\end{aligned}$
Hence, slope of the line perpendicular to line (iii)
$m=\frac{-1}{6 / 7}=\frac{-7}{6}$
$\therefore$ Equation of required line passing through $\left(\frac{-1}{17}, \frac{22}{17}\right)$ and having slope $\frac{-7}{6}$ is,
$
\begin{aligned}
y-\frac{22}{17} & =\frac{-7}{6}\left(x+\frac{1}{17}\right) \\
6(17 y-22) & =-7(17 x+1) \\
102 y-132 & =-119 x-7 \\
\Rightarrow \quad 119 x+102 y & =125
\end{aligned}
$