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Step-by-Step Solution
Step 1: Understand the Problem
We have a man starting at point P(-3,\,4) , then traveling to some point R on the x-axis (with coordinates (x,0) ), and finally reaching point Q(0,\,2) . The man walks at constant speed. To minimize the travel time, we must minimize the total distance PR + RQ . We then need to find the value of
50 \bigl( (PR)^2 + (RQ)^2 \bigr)
at this minimal distance.
Step 2: Reflect Q Across the x-axis to Find the Optimal Path
The quickest path that touches the x-axis can be found by reflecting Q(0,\,2) across the x-axis to Q'(0,\,-2) . Then, the straight line joining P(-3,\,4) and Q'(0,\,-2) will intersect the x-axis at the point R that minimizes PR + RQ .
Step 3: Equation of the Line PQ'
Let P(-3,\,4) and Q'(0,\,-2) . The slope m of line PQ' is:
m \;=\; \dfrac{-2 - 4}{0 - (-3)}
\;=\; \dfrac{-6}{3}
\;=\; -2.
Using the point-slope form with point Q'(0,\,-2) :
y - (-2) \;=\; -2 \bigl( x - 0 \bigr)
\quad\Longrightarrow\quad
y + 2 = -2x.
Hence, the equation of line PQ' is
2x + y + 2 = 0.
Step 4: Intersection of PQ' with the x-axis
On the x-axis, y=0 . Substituting in 2x + y + 2 = 0 :
2x + 0 + 2 = 0
\quad\Longrightarrow\quad
2x = -2 \quad\Longrightarrow\quad x = -1.
Therefore, R is (-1,\,0) .
Step 5: Calculate Distances PR and RQ
Distance PR :
P(-3,\,4) , R(-1,\,0) . Using the distance formula:
PR = \sqrt{((-1) - (-3))^2 + (0 - 4)^2}
= \sqrt{(2)^2 + (-4)^2}
= \sqrt{4 + 16}
= \sqrt{20}
= 2\sqrt{5}.
Distance RQ :
R(-1,\,0) , Q(0,\,2) . Again, using the distance formula:
RQ = \sqrt{(0 - (-1))^2 + (2 - 0)^2}
= \sqrt{1 + 4}
= \sqrt{5}.
Step 6: Compute (PR)^2 + (RQ)^2
(PR)^2 = (2\sqrt{5})^2 = 4 \times 5 = 20,
(RQ)^2 = (\sqrt{5})^2 = 5.
Therefore,
(PR)^2 + (RQ)^2 = 20 + 5 = 25.
Step 7: Multiply by 50
Finally,
50 \bigl( (PR)^2 + (RQ)^2 \bigr) = 50 \times 25 = 1250.
Final Answer
50 \bigl( (PR)^2 + (RQ)^2 \bigr) = 1250.
