Browse Questions

# 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.

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

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
$119x+102y=125$