Verified Solution • 2025–26 Session moderate

QJEEM25Maa347: JEE Main 2025 Mathematics - Class 11 JEEM25Maa347 Solution: Let the triangle PQR be the image of the triangle with vertices …

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+2 y=2 . If the centroid of △PQR\triangle \mathrm{PQR} is the point (α,β)(\alpha, \beta) , then 15(α−β)15(\alpha-\beta) is equal to :

Multiple Choice Options

  • A 21
  • B 19
  • C 22
  • D 24

Step-by-Step Verified Solution

Let ' GG ' be the centroid of Δ\Delta formed by (1,3)(3,1)(1,3)(3,1) \& (2,4)(2,4)

G≅(2,83)\mathrm{G} \cong\left(2, \frac{8}{3}\right)

Image of G w.r.t. x+2y−2=0x+2 y-2=0

amp;α−21=β−832=−2(2+163−2)1+4amp;=−25(163)amp;⇒α=−3215+2=−215,β=−32×215+83=−2415amp;15(α−β)=−2+24=22\begin{aligned} & \frac{\alpha-2}{1}=\frac{\beta-\frac{8}{3}}{2}=-2 \frac{\left(2+\frac{16}{3}-2\right)}{1+4} \\ & =\frac{-2}{5}\left(\frac{16}{3}\right) \\ & \Rightarrow \alpha=\frac{-32}{15}+2=\frac{-2}{15}, \beta=\frac{-32 \times 2}{15}+\frac{8}{3}=\frac{-24}{15} \\ & 15(\alpha-\beta)=-2+24=22 \end{aligned}

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