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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- 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$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...