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Straight Lines and Pair of Straight Lines question

2019 · 10 Jan · Shift 1 · Q39
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Straight Lines and Pair of Straight Lines question

2019 · 10 Jan · Shift 1 · Q39

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and R (3, – 2) are fixed points, then the locus of the centroid of Δ\DeltaΔ PQR is a line :
  1. A
    parallel to y-axis
  2. B
    with slope 23{2 \over 3}32​
  3. C
    parallel to x-axis
  4. D
    with slope 32{3 \over 2}23​
View written solutionFree

Correct answer: B

  1. Let the moving point be P(x,y)P(x,y)P(x,y) on the line 2x−3y+4=0.2x-3y+4=0.2x−3y+4=0.

  2. The fixed points are Q(1,4),R(3,−2).Q(1,4), \quad R(3,-2).Q(1,4),R(3,−2).

  3. If the centroid of △PQR\triangle PQR△PQR is G(h,k)G(h,k)G(h,k), then using the centroid formula: h=x+1+33=x+43,h=\frac{x+1+3}{3}=\frac{x+4}{3},h=3x+1+3​=3x+4​, k=y+4+(−2)3=y+23.k=\frac{y+4+(-2)}{3}=\frac{y+2}{3}.k=3y+4+(−2)​=3y+2​.

So, x=3h−4,y=3k−2.x=3h-4, \qquad y=3k-2.x=3h−4,y=3k−2.

  1. Since P(x,y)P(x,y)P(x,y) lies on 2x−3y+4=02x-3y+4=02x−3y+4=0, substitute these values: 2(3h−4)−3(3k−2)+4=0.2(3h-4)-3(3k-2)+4=0.2(3h−4)−3(3k−2)+4=0.

  2. Simplify: 6h−8−9k+6+4=06h-8-9k+6+4=06h−8−9k+6+4=0 6h−9k+2=0.6h-9k+2=0.6h−9k+2=0.

Thus, the locus of the centroid is 6h−9k+2=0.6h-9k+2=0.6h−9k+2=0.

  1. Write it in slope form: −9k=−6h−2-9k=-6h-2−9k=−6h−2 k=23h+29.k=\frac{2}{3}h+\frac{2}{9}.k=32​h+92​.

Hence the slope is 23.\frac{2}{3}.32​.

  1. Therefore, the locus is a line with slope 23\dfrac{2}{3}32​.

So the correct option is B.

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