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=

−1

17

and y=

22

17

∴ Junction point A=(

−1

17

,

22

17

)
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(

−1

17

,

22

17

)
∴ Slope of line (iii) =

−Coefficient‌of‌x

Coefficient‌of‌y

=

−6

−7

=

6

7

Hence, slope of the line perpendicular to line (iii)
m=

−1

6 7

=

−7

6

∴ Equation of required line passing through
(

−1

17

,

22

17

) and having slope

−7

6

is,
y−

22

17

=

−7

6

(x+

1

17

)
6(17y−22)=−7(17x+1)
102y−132=−119x−7
⇒ 119x+102y=125