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# 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.

Q : 24    A person standing at the junction (crossing) of two straight paths represented by the equations  $\small 2x-3y+4=0$  and  $\small 3x+4y-5=0$  wants to reach the path whose equation is  $\small 6x-7y+8=0$  in the least time. Find equation of the path that  he should follow.

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point of intersection of lines  $\small 2x-3y+4=0$  and  $\small 3x+4y-5=0$ (junction) is  $\left ( -\frac{1}{17},\frac{22}{17} \right )$
Now, person reaches to path  $\small 6x-7y+8=0$ in least time  when it follow the path perpendicular to it
Now,
Slope of line $\small 6x-7y+8=0$ is , $m'=\frac{6}{7}$
let the slope of line perpendicular to it is , m
Then,
$m= -\frac{1}{m}= -\frac{7}{6}$
Now, equation of line passing through point $\left ( -\frac{1}{17},\frac{22}{17} \right )$  and with slope $-\frac{7}{6}$  is
$\left ( y-\frac{22}{17} \right )= -\frac{7}{6}\left ( x-(-\frac{1}{17}) \right )$
$\Rightarrow 6(17y-22)=-7(17x+1)$
$\Rightarrow 102y-132=-119x-7$
$\Rightarrow 119x+102y=125$

Therefore, the required equation of line is   $119x+102y=125$

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