Let ABC 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 ABC in the line 3x + 6y – 53 = 0. Then is equal to : [2025]

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Solution

Q.
Let ABC 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 $△$ ABC in the line 3x + 6y – 53 = 0. Then $h^{2} + k^{2} + h k$ is equal to : [2025]

1 40

2 36

3 47

4 37

Ans.
(4)

Centroid of $△ A B C = (\frac{9 + 3 + 5}{3}, \frac{11 + 4 + 13}{3}) = (\frac{17}{3}, \frac{28}{3})$

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

$\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = \frac{- 2 (3 \times \frac{17}{3} + 6 \times \frac{28}{3} - 53)}{3^{2} + 6^{2}}$

Solving, we get h = 3 and k = 4

$∴ h^{2} + k^{2} + h k = 37$ .

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