AP EAPCET 25-Aug-2021 Shift 2 Paper

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 $y=\frac{22}{17}$ ∴ 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) $\text{from}\left(\frac{-1}{17},\frac{22}{17}\right)$ ∴ Slope of line (iii) $=\frac{-\text{Coefficient of }x}{\text{Coefficient of }y}=\frac{-6}{-7}=\frac{6}{7}$ Hence, slope of the line perpendicular to line (iii) $m=\frac{-1}{\frac{6}{7}}=\frac{-7}{6}$ ∴ Equation of required line passing through $\left(\frac{-1}{17},\frac{22}{17}\right)$ and having slope $\frac{-7}{6}$ is, $y-\frac{22}{17}=\frac{-7}{6}\left(x+\frac{1}{17}\right)$ $6(17y-22)=-7(17x+1)$ $102y-132=-119x-7$ ⇒ $119x+102y=125$