A person standing at the junction of two stra | vedclass

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(A) The equations of the given lines are:$2x - 3y + 4 = 0$ $(1)$$3x + 4y - 5 = 0$ $(2)$$6x - 7y + 8 = 0$ $(3)$The person is standing at the junction o

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$119x + 102y = 125$

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(A) The equations of the given lines are:$2x - 3y + 4 = 0$ $(1)$$3x + 4y - 5 = 0$ $(2)$$6x - 7y + 8 = 0$ $(3)$The person is standing at the junction of the paths represented by lines $(1)$ and $(2)$.Solving equations $(1)$ and $(2)$ by multiplying $(1)$ by $4$ and $(2)$ by $3$:$8x - 12y + 16 = 0$$9x + 12y - 15 = 0$Adding these,we get $17x + 1 = 0$,so $x = -\frac{1}{17}$.Substituting $x$ into $(1)$: $2(-\frac{1}{17}) - 3y + 4 = 0 \implies -\frac{2}{17} + 4 = 3y \implies \frac{66}{17} = 3y \implies y = \frac{22}{17}$.Thus,the person is at point $(-\frac{1}{17}, \frac{22}{17})$.To reach path $(3)$ in the least time,the person must walk along the perpendicular line to $(3)$ from the point $(-\frac{1}{17}, \frac{22}{17})$.The slope of line $(3)$ is $m = \frac{6}{7}$.The slope of the perpendicular line is $m' = -\frac{1}{m} = -\frac{7}{6}$.The equation of the line passing through $(-\frac{1}{17}, \frac{22}{17})$ with slope $-\frac{7}{6}$ is:$(y - \frac{22}{17}) = -\frac{7}{6}(x + \frac{1}{17})$$6(17y - 22) = -7(17x + 1)$$102y - 132 = -119x - 7$$119x + 102y = 125$.

$A$ person standing at the junction 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 the equation of the path that he should follow.

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