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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.
 

Answers (1)

best_answer

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

Posted by

Gautam harsolia

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