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Straight Lines and Pair of Straight Lines question

2021 · 1 Sep · Shift 2 · Q46
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Straight Lines and Pair of Straight Lines question

2021 · 1 Sep · Shift 2 · Q46

JEE MainMathematicsStraight Lines and Pair of Straight LinesNumerical+4 / −1
A man starts walking from the point P(−-− 3, 4), touches the x-axis at R, and then turns to reach at the point Q(0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then 50((PR)2+(RQ)2)50\left( {{{(PR)}^2} + {{(RQ)}^2}} \right)50((PR)2+(RQ)2) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 1250

  1. Since the man walks at constant speed, minimizing time is equivalent to minimizing the total distance PR+RQPR+RQPR+RQ where RRR lies on the xxx-axis.

  2. Use the reflection principle.

    Reflect Q(0,2)Q(0,2)Q(0,2) in the xxx-axis. Its image is Q′(0,−2).Q'(0,-2).Q′(0,−2).

    For a path from P(−3,4)P(-3,4)P(−3,4) to the xxx-axis and then to Q(0,2)Q(0,2)Q(0,2), the minimum value of PR+RQPR+RQPR+RQ occurs when the path is equivalent to the straight line from PPP to Q′Q'Q′ crossing the xxx-axis at RRR.

  3. Find the line joining P(−3,4)P(-3,4)P(−3,4) and Q′(0,−2)Q'(0,-2)Q′(0,−2).

    Slope: m=−2−40−(−3)=−63=−2.m=\frac{-2-4}{0-(-3)}=\frac{-6}{3}=-2.m=0−(−3)−2−4​=3−6​=−2.

    Equation through P(−3,4)P(-3,4)P(−3,4): y−4=−2(x+3)y-4=-2(x+3)y−4=−2(x+3) y=−2x−2.y=-2x-2.y=−2x−2.

  4. Since RRR lies on the xxx-axis, at RRR we have y=0y=0y=0.

    So, 0=−2x−2  ⟹  x=−1.0=-2x-2 \implies x=-1.0=−2x−2⟹x=−1.

    Hence, R=(−1,0).R=(-1,0).R=(−1,0).

  5. Now compute PR2PR^2PR2 and RQ2RQ^2RQ2.

    PR2=[(−1)−(−3)]2+(0−4)2=22+(−4)2=4+16=20.PR^2 = [(-1)-(-3)]^2+(0-4)^2 = 2^2+(-4)^2 = 4+16=20.PR2=[(−1)−(−3)]2+(0−4)2=22+(−4)2=4+16=20.

    RQ2=[0−(−1)]2+(2−0)2=12+22=1+4=5.RQ^2 = [0-(-1)]^2+(2-0)^2 = 1^2+2^2 = 1+4=5.RQ2=[0−(−1)]2+(2−0)2=12+22=1+4=5.

  6. Therefore, PR2+RQ2=20+5=25.PR^2+RQ^2=20+5=25.PR2+RQ2=20+5=25.

  7. Required value: 50(PR2+RQ2)=50(25)=1250.50\left(PR^2+RQ^2\right)=50(25)=1250.50(PR2+RQ2)=50(25)=1250.

  8. Final answer: 1250\boxed{1250}1250​

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