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MCQMediumJEE 2025Basics: Distance, Section Formula, Locus

JEE Mathematics 2025 Question with Solution

Let the triangle PQR be the image of the triangle with vertices (1,3),(3,1)(1, 3), (3, 1) and (2,4)(2, 4) in the line x+2y=2x + 2y = 2. If the centroid of ΔPQR\Delta PQR is the point (α,β)(\alpha, \beta), then 15(α−β)15(\alpha - \beta) is equal to :

  • A

    2424

  • B

    1919

  • C

    2121

  • D

    2222

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The triangle with vertices (1,3),(3,1),(2,4)(1,3), (3,1), (2,4) is reflected in the line x+2y−2=0x+2y-2=0 to form triangle PQRPQR.

Find: If the centroid of ΔPQR\Delta PQR is (α,β)(\alpha, \beta), find 15(α−β)15(\alpha-\beta).

The reflection of a point (x0,y0)(x_0,y_0) in the line ax+by+c=0ax+by+c=0 is given by

(x′,y′)=(x0−2a(ax0+by0+c)a2+b2,  y0−2b(ax0+by0+c)a2+b2).(x',y')=\left(x_0-\frac{2a(ax_0+by_0+c)}{a^2+b^2},\; y_0-\frac{2b(ax_0+by_0+c)}{a^2+b^2}\right).

Here a=1,  b=2,  c=−2a=1,\; b=2,\; c=-2, so a2+b2=5a^2+b^2=5.

Reflect (1,3)(1,3):

ax+by+c=1+2⋅3−2=5ax+by+c = 1+2\cdot 3-2 = 5

So,

x′=1−2⋅1⋅55=−1,y′=3−2⋅2⋅55=−1.x' = 1-\frac{2\cdot 1\cdot 5}{5} = -1, \qquad y' = 3-\frac{2\cdot 2\cdot 5}{5} = -1.

Hence the image is (−1,−1)(-1,-1).

Reflect (3,1)(3,1):

ax+by+c=3+2⋅1−2=3ax+by+c = 3+2\cdot 1-2 = 3

So,

x′=3−2⋅1⋅35=95,y′=1−2⋅2⋅35=−75.x' = 3-\frac{2\cdot 1\cdot 3}{5} = \frac{9}{5}, \qquad y' = 1-\frac{2\cdot 2\cdot 3}{5} = -\frac{7}{5}.

Hence the image is (95,−75)\left(\frac{9}{5},-\frac{7}{5}\right).

Reflect (2,4)(2,4):

ax+by+c=2+2⋅4−2=8ax+by+c = 2+2\cdot 4-2 = 8

So,

x′=2−2⋅1⋅85=−65,y′=4−2⋅2⋅85=−125.x' = 2-\frac{2\cdot 1\cdot 8}{5} = -\frac{6}{5}, \qquad y' = 4-\frac{2\cdot 2\cdot 8}{5} = -\frac{12}{5}.

Hence the image is (−65,−125)\left(-\frac{6}{5},-\frac{12}{5}\right).

Therefore, the reflected triangle has vertices

P(−1,−1),Q(95,−75),R(−65,−125).P(-1,-1),\quad Q\left(\frac{9}{5},-\frac{7}{5}\right),\quad R\left(-\frac{6}{5},-\frac{12}{5}\right).

Now the centroid (α,β)(\alpha,\beta) is the average of the coordinates:

α=−1+95−653,β=−1−75−1253.\alpha = \frac{-1+\frac{9}{5}-\frac{6}{5}}{3}, \qquad \beta = \frac{-1-\frac{7}{5}-\frac{12}{5}}{3}.

Simplifying,

−1+95−65=−55+35=−25-1+\frac{9}{5}-\frac{6}{5} = -\frac{5}{5}+\frac{3}{5} = -\frac{2}{5}

so

α=−2/53=−215.\alpha = \frac{-2/5}{3} = -\frac{2}{15}.

Also,

−1−75−125=−55−195=−245-1-\frac{7}{5}-\frac{12}{5} = -\frac{5}{5}-\frac{19}{5} = -\frac{24}{5}

so

β=−24/53=−85.\beta = \frac{-24/5}{3} = -\frac{8}{5}.

Now,

α−β=−215−(−85)=−215+2415=2215.\alpha-\beta = -\frac{2}{15} - \left(-\frac{8}{5}\right) = -\frac{2}{15}+\frac{24}{15}=\frac{22}{15}.

Hence,

15(α−β)=15⋅2215=22.15(\alpha-\beta)=15\cdot \frac{22}{15}=22.

Therefore, the correct option is D.

Triangle with vertices marked (1,3), (3,1), and (2,4), centroid shown, line x+2y-2=0 below, and reflected centroid labeled alpha beta with calculation steps.

Common mistakes

  • Using the centroid of the original triangle as the final answer. That is wrong because the triangle is reflected first, so the centroid must also be reflected or recomputed from the reflected vertices. First transform the triangle, then find (α,β)(\alpha,\beta).

  • Applying the reflection formula with the wrong sign of cc in the line equation. The line must be written as x+2y−2=0x+2y-2=0, so c=−2c=-2, not 22. Using the wrong value changes every reflected coordinate.

  • Averaging the reflected coordinates incorrectly. The centroid is found by dividing the sum of the three xx-coordinates and three yy-coordinates by 33. Do not divide intermediate partial sums inconsistently.

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