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Ncert Solutions Maths class 11th Maths Class 11th Straight Lines

76. 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.

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alok kumar singh | Contributor-Level 10

4 months ago

76.

If point P be the junction between the lines

2x – 3y + 4 = 0 ______ (1)

3x + 4y – 5 = 0 ______ (2)

Solving (1) and (2) using 3 × (1) – 2 × (2) we get,

6x – 9y + 12 – (6x + 8y – 10) = 0

–17y + 22 = 0

y = $\frac{22}{17}$

And 2x = 3y– 4

=> 2x = 3 × $\frac{22}{17}$ – 4

x = $\frac{33}{17}$ – 2 = $\frac{33 - 34}{17}$ = $- \frac{1}{17}$

Hence, the co-ordinate of the junction is P $(- \frac{1}{17}, \frac{22}{17})$

The eqⁿ of the path to be reach is

6x – 7y + 8 = 0 _____ (3)

Then, least distance will be perpendicular path.

So, slope of ⊥ path = $- 1$
$slope of line (3)$

$= \frac{- 1}{(6 / 7)} = \frac{- 7}{6}$

Hence eqⁿ of shortest/least distance path from P $(- \frac{1}{17}, \frac{22}{17})$ is

$y - \frac{22}{17} = \frac{- 7}{6} (x + \frac{1}{17}) .$

$\Rightarrow 6 y - \frac{132}{17} = - 7 x - \frac{7}{17} .$

$\Rightarrow 7 x + 6 y - \frac{132}{17} + \frac{7}{17} = 0$

$\Rightarrow 7 x + 6 y - \frac{125}{17} = 0 .$

119x + 102y &nd

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76.

If point P be the junction between the lines

2x – 3y + 4 = 0 ______ (1)

3x + 4y – 5 = 0 ______ (2)

Solving (1) and (2) using 3 × (1) – 2 × (2) we get,

6x – 9y + 12 – (6x + 8y – 10) = 0

–17y + 22 = 0

y = $\frac{22}{17}$

And 2x = 3y– 4

=> 2x = 3 × $\frac{22}{17}$ – 4

x = $\frac{33}{17}$ – 2 = $\frac{33 - 34}{17}$ = $- \frac{1}{17}$

Hence, the co-ordinate of the junction is P $(- \frac{1}{17}, \frac{22}{17})$

The eqⁿ of the path to be reach is

6x – 7y + 8 = 0 _____ (3)

Then, least distance will be perpendicular path.

So, slope of ⊥ path = $- 1$
$slope of line (3)$

$= \frac{- 1}{(6 / 7)} = \frac{- 7}{6}$

Hence eqⁿ of shortest/least distance path from P $(- \frac{1}{17}, \frac{22}{17})$ is

$y - \frac{22}{17} = \frac{- 7}{6} (x + \frac{1}{17}) .$

$\Rightarrow 6 y - \frac{132}{17} = - 7 x - \frac{7}{17} .$

$\Rightarrow 7 x + 6 y - \frac{132}{17} + \frac{7}{17} = 0$

$\Rightarrow 7 x + 6 y - \frac{125}{17} = 0 .$

119x + 102y – 125 = 0

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76. 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.

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