Verified Solution • 2025–26 Session moderate

QJEEM25Ma3a16: JEE Main 2025 Mathematics - Class 11 JEEM25Ma3a16 Solution: Let ΔABC be a triangle formed by the lines 7x – 6y + 3 = 0, x + …

Let ΔABC be a triangle formed by the lines 7x – 6y + 3 = 0, x + 2y – 31 = 0 and 9x – 2y – 19 = 0. Let the point (h, k) be the image of the centroid of ΔABC in the line 3x + 6y – 53 = 0. Then h2 + k2 + hk is equal to :

Multiple Choice Options

  • A 47
  • B 37
  • C 40
  • D 36

Step-by-Step Verified Solution

JEE Main 2025 (Online) 29th January Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 9 English Explanation 1

∴ centroid of △ABCamp;=(9+3+53,11+4+133)amp;=(173,283)\begin{aligned}\therefore \text { centroid of } \triangle \mathrm{ABC} & =\left(\frac{9+3+5}{3}, \frac{11+4+13}{3}\right) \\& =\left(\frac{17}{3}, \frac{28}{3}\right)\end{aligned}

JEE Main 2025 (Online) 29th January Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 9 English Explanation 2

Let image of centroid with respect to line mirror is (h, k)

amp;amp;∴(k−283 h−173)(−12)=−1amp;&3( h+1732)+6(k+2832)=53amp; Solving (1) & (2) we get h=3,k=4amp;∴h2+k2+hk=37\begin{aligned}&\begin{aligned}& \therefore\left(\frac{\mathrm{k}-\frac{28}{3}}{\mathrm{~h}-\frac{17}{3}}\right)\left(-\frac{1}{2}\right)=-1 \\& \& 3\left(\frac{\mathrm{~h}+\frac{17}{3}}{2}\right)+6\left(\frac{\mathrm{k}+\frac{28}{3}}{2}\right)=53\end{aligned}\\&\text { Solving (1) \& (2) we get } \mathrm{h}=3, \mathrm{k}=4\\&\therefore \mathrm{h}^2+\mathrm{k}^2+\mathrm{hk}=37\end{aligned}

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