Starting from the point A(-3, 4),a moving obj | vedclass

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Question

(A) To find the shortest path from $A(-3, 4)$ to $C(0, 1)$ touching the line $L: 2x + y - 7 = 0$,we reflect point $A$ across the line $L$ to get $A'$.

Vedclass · Accepted Answer

$\frac{9 \sqrt{170}}{25}$

Vedclass · Answer

(A) To find the shortest path from $A(-3, 4)$ to $C(0, 1)$ touching the line $L: 2x + y - 7 = 0$,we reflect point $A$ across the line $L$ to get $A'$.Let $A'(x', y')$ be the reflection of $A(-3, 4)$ in $2x + y - 7 = 0$.Using the formula $\frac{x' - (-3)}{2} = \frac{y' - 4}{1} = -2 \frac{2(-3) + 4 - 7}{2^2 + 1^2} = -2 \frac{-9}{5} = \frac{18}{5}$.$x' = -3 + 2(\frac{18}{5}) = -3 + \frac{36}{5} = \frac{21}{5}$.$y' = 4 + 1(\frac{18}{5}) = 4 + \frac{18}{5} = \frac{38}{5}$.So,$A' = (\frac{21}{5}, \frac{38}{5})$.The shortest path distance is the length of the line segment $A'C$.The point $B$ is the intersection of $A'C$ and the line $2x + y - 7 = 0$.The line $A'C$ passes through $(\frac{21}{5}, \frac{38}{5})$ and $(0, 1)$.Slope $m = \frac{1 - 38/5}{0 - 21/5} = \frac{-33/5}{-21/5} = \frac{33}{21} = \frac{11}{7}$.Equation of $A'C$: $y - 1 = \frac{11}{7}(x - 0) \Rightarrow 11x - 7y + 7 = 0$.Solving $2x + y - 7 = 0$ and $11x - 7y + 7 = 0$:$y = 7 - 2x$ $\Rightarrow 11x - 7(7 - 2x) + 7 = 0$ $\Rightarrow 11x - 49 + 14x + 7 = 0$ $\Rightarrow 25x = 42$ $\Rightarrow x = \frac{42}{25}$.$y = 7 - 2(\frac{42}{25}) = \frac{175 - 84}{25} = \frac{91}{25}$.So,$B = (\frac{42}{25}, \frac{91}{25})$.Distance $AB = \sqrt{(\frac{42}{25} + 3)^2 + (\frac{91}{25} - 4)^2} = \sqrt{(\frac{117}{25})^2 + (\frac{-9}{25})^2} = \frac{1}{25} \sqrt{13689 + 81} = \frac{\sqrt{13770}}{25} = \frac{9 \sqrt{170}}{25}$.

Starting from the point $A(-3, 4)$,a moving object touches the line $2x + y - 7 = 0$ at point $B$ and reaches the point $C(0, 1)$. If the object travels along the shortest path,the distance between $A$ and $B$ is:

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