A person standing at the junction (crossing) | vedclass

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(B) The person is standing at the intersection of the lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$. Solving these equations,we multiply the first by

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$119x + 102y - 125 = 0$

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(B) The person is standing at the intersection of the lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$. Solving these equations,we multiply the first by $4$ and the second by $3$: $8x - 12y + 16 = 0$ $9x + 12y - 15 = 0$ Adding these gives $17x + 1 = 0$,so $x = -\frac{1}{17}$. Substituting $x$ into the first equation: $2(-\frac{1}{17}) - 3y + 4 = 0$ $\Rightarrow -\frac{2}{17} + 4 = 3y$ $\Rightarrow 3y = \frac{66}{17}$ $\Rightarrow y = \frac{22}{17}$. The junction point is $A = (-\frac{1}{17}, \frac{22}{17})$. To reach the path $6x - 7y + 8 = 0$ in the least time,the person must walk along the perpendicular line. The slope of the given path is $m_1 = \frac{6}{7}$. The slope of the perpendicular line is $m_2 = -\frac{1}{m_1} = -\frac{7}{6}$. The equation of the line passing through $(-\frac{1}{17}, \frac{22}{17})$ with slope $-\frac{7}{6}$ is: $y - \frac{22}{17} = -\frac{7}{6}(x + \frac{1}{17})$ $6(17y - 22) = -7(17x + 1)$ $102y - 132 = -119x - 7$ $119x + 102y - 125 = 0$.

$A$ person standing at the junction (crossing) of $2$ 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. The equation of the path he should follow is:

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