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

• A. 24
• B. 19
• C. 21
• D. 22

Hint:

The centroid of a triangle is the average of the coordinates of its vertices.

Answer 1

The Correct Option is D

Let \( G \) be the centroid of \( \Delta PQR \) formed by \( (1, 3), (3, 1), (2, 4) \). The centroid of a triangle is the average of the coordinates of its vertices: \[ G = \left( \frac{1+3+2}{3}, \frac{3+1+4}{3} \right) = \left( \frac{6}{3}, \frac{8}{3} \right) = \left( 2, \frac{8}{3} \right). \] Next, we find the image of \( G \) under the transformation \( x + 2y = 2 \). The transformation matrix for this is: \[ \alpha - 2 = \frac{-2}{5}, \quad \beta - \frac{8}{3} = \frac{-32}{15} + 2, \] which leads to \( \alpha = -\frac{2}{5} \) and \( \beta = -\frac{24}{15} \). Thus, \( 15(\alpha - \beta) = 15(-\frac{2}{5} - \frac{24}{15}) = 22 \). Thus, \( 15(\alpha - \beta) = \boxed{22} \).

Answer 2

Answer 3

1. Write the line in the form $ax+by+c=0$: here $x+2y-2=0$, so $a=1,\;b=2,\;c=-2$.
2. Formula for reflection of a point $(x_0,y_0)$ in the line $ax+by+c=0$ is $$ (x',y')=\Big(x_0-\dfrac{2a(ax_0+by_0+c)}{a^2+b^2},\; y_0-\dfrac{2b(ax_0+by_0+c)}{a^2+b^2}\Big). $$
3. Reflect each vertex.
   • For $(1,3)$: compute $ax+by+c = 1\cdot1 + 2\cdot3 -2 = 1+6-2=5$. $a^2+b^2=1^2+2^2=5$. Thus $$x' = 1 - \frac{2\cdot1\cdot5}{5} = 1-2 = -1,\qquad y' = 3 - \frac{2\cdot2\cdot5}{5} = 3-4 = -1.$$ So image is $(-1,-1)$.
   • For $(3,1)$: $ax+by+c = 3 + 2\cdot1 -2 = 3+2-2=3$. Then $$x' = 3 - \frac{2\cdot1\cdot3}{5} = 3 - \frac{6}{5} = \frac{9}{5},$$ $$y' = 1 - \frac{2\cdot2\cdot3}{5} = 1 - \frac{12}{5} = -\frac{7}{5}.$$ So image is $\big(\tfrac{9}{5},-\tfrac{7}{5}\big)$.
   • For $(2,4)$: $ax+by+c = 2 + 2\cdot4 -2 = 2+8-2=8$. Then $$x' = 2 - \frac{2\cdot1\cdot8}{5} = 2 - \frac{16}{5} = -\frac{6}{5},$$ $$y' = 4 - \frac{2\cdot2\cdot8}{5} = 4 - \frac{32}{5} = -\frac{12}{5}.$$ So image is $\big(-\tfrac{6}{5},-\tfrac{12}{5}\big)$.
4. The reflected triangle $PQR$ has vertices $$P(-1,-1),\quad Q\Big(\tfrac{9}{5},-\tfrac{7}{5}\Big),\quad R\Big(-\tfrac{6}{5},-\tfrac{12}{5}\Big).$$
5. Centroid $(\alpha,\beta)$ is the average of coordinates: $$ \alpha=\frac{-1+\tfrac{9}{5}-\tfrac{6}{5}}{3},\qquad \beta=\frac{-1-\tfrac{7}{5}-\tfrac{12}{5}}{3}. $$ Simplify $\alpha$: $$-1+\tfrac{9}{5}-\tfrac{6}{5} = -\tfrac{5}{5}+\tfrac{3}{5} = -\tfrac{2}{5},\quad \alpha=\frac{-2/5}{3}=-\frac{2}{15}.$$ Simplify $\beta$: $$-1-\tfrac{7}{5}-\tfrac{12}{5} = -\tfrac{5}{5}-\tfrac{19}{5} = -\tfrac{24}{5},\quad \beta=\frac{-24/5}{3}=-\frac{8}{5}.$$
6. Compute $$\alpha-\beta = -\frac{2}{15} -\Big(-\frac{8}{5}\Big) = -\frac{2}{15} + \frac{8}{5} = -\frac{2}{15} + \frac{24}{15} = \frac{22}{15}.$$ Hence $$15(\alpha-\beta)=15\cdot\frac{22}{15}=\boxed{22}.$$

Answer 

$15(\alpha-\beta)=22$. (Option 4)