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Straight Lines and Pair of Straight Lines question

2004 · Shift 0 · Q108
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Straight Lines and Pair of Straight Lines question

2004 · Shift 0 · Q108

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let A(2,−3)A\left( {2, - 3} \right)A(2,−3) and B(−2,1)B\left( {-2, 1} \right)B(−2,1) be vertices of a triangle ABCABCABC. If the centroid of this triangle moves on the line 2x+3y=12x + 3y = 12x+3y=1, then the locus of the vertex CCC is the line :
  1. A
    3x−2y=33x - 2y = 33x−2y=3
  2. B
    2x−3y=72x - 3y = 72x−3y=7
  3. C
    3x+2y=53x + 2y = 53x+2y=5
  4. D
    2x+3y=92x + 3y = 92x+3y=9
View written solutionFree

Correct answer: D

  1. Let the coordinates of the third vertex be C(x,y).C(x,y).C(x,y).

  2. The centroid GGG of triangle ABCABCABC is given by G(xA+xB+x3,yA+yB+y3).G\left(\frac{x_A+x_B+x}{3},\frac{y_A+y_B+y}{3}\right).G(3xA​+xB​+x​,3yA​+yB​+y​).

    Here, A(2,−3),B(−2,1).A(2,-3),\quad B(-2,1).A(2,−3),B(−2,1).

    So, xA+xB+x=2+(−2)+x=x,x_A+x_B+x=2+(-2)+x=x,xA​+xB​+x=2+(−2)+x=x, yA+yB+y=−3+1+y=y−2.y_A+y_B+y=-3+1+y=y-2.yA​+yB​+y=−3+1+y=y−2.

    Hence the centroid is G(x3,y−23).G\left(\frac{x}{3},\frac{y-2}{3}\right).G(3x​,3y−2​).

  3. It is given that the centroid moves on the line 2X+3Y=1,2X+3Y=1,2X+3Y=1, where G=(X,Y)G=(X,Y)G=(X,Y).

    Substitute X=x3,Y=y−23.X=\frac{x}{3},\quad Y=\frac{y-2}{3}.X=3x​,Y=3y−2​.

    Then, 2(x3)+3(y−23)=1.2\left(\frac{x}{3}\right)+3\left(\frac{y-2}{3}\right)=1.2(3x​)+3(3y−2​)=1.

  4. Simplify: 2x3+y−2=1\frac{2x}{3}+y-2=132x​+y−2=1 2x3+y=3\frac{2x}{3}+y=332x​+y=3 Multiply by 333: 2x+3y=9.2x+3y=9.2x+3y=9.

  5. Therefore, the locus of vertex CCC is 2x+3y=9.\boxed{2x+3y=9}. 2x+3y=9​.

  6. Checking options:

    • A: 3x−2y=33x-2y=33x−2y=3 ❌
    • B: 2x−3y=72x-3y=72x−3y=7 ❌
    • C: 3x+2y=53x+2y=53x+2y=5 ❌
    • D: 2x+3y=92x+3y=92x+3y=9 ✅

Therefore, the correct option is D.

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