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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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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. Find equation of the path that he should follow.

$\begin {array} {1 1} (A)\;119x-102y=125 & \quad (B)\;-119x+102y=125 \\ (C)\;119x+102y=-125 & \quad (D)\;119x+102y=125 \end {array}$

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Toolbox:
  • Slope of the line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is $ m = \large\frac{y_2-y_1}{x_2-x_1}$
  • Equation of a line passing through the point $(x_1, y_1)$ and having slope $m$ is $ y-y_1 = m(x-x_1)$
The given equations are :
$ \qquad 2x-3y+4=0$------(1)
$ \qquad 3x+4y-5=0$--------(2)
$ \qquad 6x-7y+8=0$--------(3)
It is given that the person is standing at the junction of the paths represented by equation (1) and (2).
Solving equation (1) and (2) for $x$ and $y$ we get,
$ ( \times 3) 2x-3y=-4$
$ ( \times 2) 3x+4y=5$
$ \qquad 6x-9y=-12$
$ \qquad 6x + 8y=10$
$ \qquad \qquad -17y=-22$
$ \qquad \qquad y = \large\frac{22}{17}$
$ \therefore 2x-3 \bigg( \large\frac{22}{17} \bigg)$$ = -4$
$ \Rightarrow 2x=-4 + \large\frac{66}{17}$
$ \therefore x = -\large\frac{1}{17}$
Hence the person is standing at point $ \bigg( \large\frac{-1}{17}$$, \large\frac{22}{17} \bigg)$
Slope of the line (3) = $\large\frac{6}{7}$
$ \therefore $ slope of the line perpendicular to line (3) is $ \large\frac{-1}{\Large\frac{6}{7}}$$ = -\large\frac{7}{6}$
Hence the equation of the line passing through $ \bigg( -\large\frac{1}{17}$$, \large\frac{22}{17} \bigg)$ and slope $ -\large\frac{7}{6}$ is
$ \bigg( y-\large\frac{22}{17} \bigg) = -\large\frac{7}{6} $$ \bigg[ x - \bigg( -\large\frac{1}{17} \bigg) \bigg]$
$ \Rightarrow y - \large\frac{22}{17} $$ =- \large\frac{7}{6}$$ \bigg( x + \large\frac{1}{17} \bigg)$
$ \Rightarrow \large\frac{17y-22}{17}$$ = -\large\frac{7}{6}$$ \large\frac{(17x+1)}{17}$
$ \Rightarrow 6(17y-22) = -7 (17x+1)$
$ 102y-132 = -119x-7$
$ \Rightarrow 119x+102y=125$
Hence the path that the person should follow
$ 119x+102y=125$
answered May 23, 2014 by thanvigandhi_1
 

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