JEE MAIN - Mathematics (2025 - 29th January Morning Shift - No. 7)
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 :
47
37
40
36
Explanation
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$$\begin{aligned} \therefore \text { centroid of } \triangle \mathrm{ABC} & =\left(\frac{9+3+5}{3}, \frac{11+4+13}{3}\right) \\ & =\left(\frac{17}{3}, \frac{28}{3}\right) \end{aligned}$$
_29th_January_Morning_Shift_en_7_2.png)
Let image of centroid with respect to line mirror is (h, k)
$$\begin{aligned} &\begin{aligned} & \therefore\left(\frac{\mathrm{k}-\frac{28}{3}}{\mathrm{~h}-\frac{17}{3}}\right)\left(-\frac{1}{2}\right)=-1 \\ & \& 3\left(\frac{\mathrm{~h}+\frac{17}{3}}{2}\right)+6\left(\frac{\mathrm{k}+\frac{28}{3}}{2}\right)=53 \end{aligned}\\ &\text { Solving (1) \& (2) we get } \mathrm{h}=3, \mathrm{k}=4\\ &\therefore \mathrm{h}^2+\mathrm{k}^2+\mathrm{hk}=37 \end{aligned}$$
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