Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line x + 2y = 2. If the centroid of PQR is the point , then is equal to: [2025]

Question

Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line x + 2y = 2. If the centroid of $△$ PQR is the point $(α, β)$ , then $15 (α - β)$ is equal to: [2025]

Smart Achievers · Accepted Answer

(4)

Let P, Q and R be the mirror images of points (1, 3), (3, 1) and (2, 4) in the line x + 2y – 2 = 0.

Coordinates of P is given by

$\frac{x - 1}{1} = \frac{y - 3}{2} = - 2 (\frac{1 + 6 - 2}{5}) i . e ., P (- 1, - 1)$

Similarly, $Q (\frac{9}{5}, \frac{- 7}{5}) and R (\frac{- 6}{5}, \frac{- 12}{5})$

Centroid of $△ P Q R = (\frac{- 2}{15}, - \frac{8}{5})$

$∴ α = \frac{- 2}{15} and β = - \frac{- 8}{5}$

Thus, $15 (α - β) = 15 (\frac{- 2}{15} + \frac{8}{5}) = 22$ .

Solution

Q.
Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line x + 2y = 2. If the centroid of $△$ PQR is the point $(α, β)$ , then $15 (α - β)$ is equal to: [2025]

1 19

2 24

3 21

4 22

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