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 $\triangle PQR$ is the point $(\alpha, \beta)$, then $15(\alpha-\beta)$ is equal to :

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Straight Lines

Let the triangle PQR be the image of the triangle with vertices

(1, 3), (3, 1)

and

(2, 4)

in the line

x + 2 y = 2

. If the centroid of

△ P QR

is the point

(α, β)

, then

15 (α - β)

is equal to :

Answer: (D)

The centroid of the original triangle with vertices

(1, 3), (3, 1), (2, 4)

G = (\frac{1 + 3 + 2}{3}, \frac{3 + 1 + 4}{3}) = (2, \frac{8}{3}) .

Reflection is an affine (isometric) map, so it commutes with taking centroids: the centroid of the reflected triangle

P QR

is just the reflection of

G

in the line

x + 2 y = 2

, i.e.

x + 2 y - 2 = 0

.

For a point

(x_{1}, y_{1})

, its image

(α, β)

in the line

a x + b y + c = 0

satisfies

\frac{α - x _{1}}{a} = \frac{β - y _{1}}{b} = \frac{- 2 ( a x _{1} + b y _{1} + c )}{a ^{2} + b ^{2}} .

Here

a = 1, b = 2, c = - 2

, and at

G = (2, 8/3)

a x_{1} + b y_{1} + c = 2 + \frac{16}{3} - 2 = \frac{16}{3}, a^{2} + b^{2} = 5.

\frac{α - 2}{1} = \frac{β - \frac{8}{3}}{2} = \frac{- 2}{5} \cdot \frac{16}{3} = - \frac{32}{15} .

α = 2 - \frac{32}{15} = - \frac{2}{15}, β = \frac{8}{3} - \frac{64}{15} = \frac{40 - 64}{15} = - \frac{24}{15} .

Then

15 (α - β) = 15 (- \frac{2}{15} + \frac{24}{15}) = 22.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option