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NCERT Class XI Mathematics - Straight Lines - Solutions

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Question : 77
Total: 74
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. Find equation of the path that he should follow.
Solution:  

The given equation of lines are
2x – 3y + 4 = 0 ...(i)
and 3x + 4y – 5 = 0 ... (ii)
On solving (i) and (ii), we get x = −
1
17
 , y =
22
17
∴ Required point is (−
1
17
,
22
17
)
Now slope of 6x – 7y + 8 = 0 is
6
7
∴ Required slope is −
7
6
Therefore required equation of a line is (y−
22
17
) = −
7
6
(x+
1
17
)
⇒ (17y – 22) 6 = – 7(17x + 1)
⇒ 102y – 132 = – 119x – 7
⇒ 119x + 102y – 125 = 0
⇒ 119x + 102y = 125.
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