Key Concepts and Formulas

• Centroid of a Triangle: For a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the centroid $G$ is given by: $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$
• Image of a Point in a Line: If a point $P(x_1, y_1)$ has an image $P'(h, k)$ in the line $Ax + By + C = 0$, then the coordinates $(h, k)$ can be found using the formula: $\frac{h-x_1}{A} = \frac{k-y_1}{B} = -2\frac{Ax_1+By_1+C}{A^2+B^2}$
• Intersection of Two Lines: Solving a system of two linear equations to find the point where two lines meet.

Step-by-Step Solution

Step 1: Determine the Vertices of $\Delta ABC$

The vertices of the triangle are the intersection points of the lines. Let's label the lines:

• Line L1: $7x - 6y + 3 = 0$
• Line L2: $x + 2y - 31 = 0$
• Line L3: $9x - 2y - 19 = 0$

We will find the intersection points for each pair of lines.

• Intersection of L1 and L2 (Vertex A): We have the system of equations:

  1. $7x - 6y = -3$
  2. $x + 2y = 31$ Multiply equation (2) by 3 to eliminate $y$: $3(x + 2y) = 3(31) \Rightarrow 3x + 6y = 93$ Now, add this modified equation to equation (1): $(7x - 6y) + (3x + 6y) = -3 + 93$ $10x = 90 \Rightarrow x = 9$ Substitute $x=9$ into equation (2): $9 + 2y = 31 \Rightarrow 2y = 22 \Rightarrow y = 11$ So, Vertex A is $(9, 11)$.

• Intersection of L2 and L3 (Vertex B): We have the system of equations: 2. $x + 2y = 31$ 3. $9x - 2y = 19$ Add equation (2) and equation (3) to eliminate $y$: $(x + 2y) + (9x - 2y) = 31 + 19$ $10x = 50 \Rightarrow x = 5$ Substitute $x=5$ into equation (2): $5 + 2y = 31 \Rightarrow 2y = 26 \Rightarrow y = 13$ So, Vertex B is $(5, 13)$.

• Intersection of L1 and L3 (Vertex C): We have the system of equations:

  1. $7x - 6y = -3$
  2. $9x - 2y = 19$ Multiply equation (3) by 3 to eliminate $y$: $3(9x - 2y) = 3(19) \Rightarrow 27x - 6y = 57$ Now, subtract equation (1) from this modified equation: $(27x - 6y) - (7x - 6y) = 57 - (-3)$ $20x = 60 \Rightarrow x = 3$ Substitute $x=3$ into equation (3): $9(3) - 2y = 19 \Rightarrow 27 - 2y = 19 \Rightarrow 2y = 8 \Rightarrow y = 4$ So, Vertex C is $(3, 4)$.

The vertices of $\Delta ABC$ are $A(9, 11)$, $B(5, 13)$, and $C(3, 4)$.

Step 2: Calculate the Centroid of $\Delta ABC$

Using the centroid formula $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$ with the vertices $A(9, 11)$, $B(5, 13)$, and $C(3, 4)$:

$G_x = \frac{9+5+3}{3} = \frac{17}{3}$ $G_y = \frac{11+13+4}{3} = \frac{28}{3}$

The centroid of $\Delta ABC$ is $G\left(\frac{17}{3}, \frac{28}{3}\right)$.

Step 3: Find the Image of the Centroid in the Given Line

Let the centroid be $P(x_1, y_1) = \left(\frac{17}{3}, \frac{28}{3}\right)$. Let the line be $L: 3x + 6y - 53 = 0$. Comparing this with the general form $Ax + By + C = 0$, we have $A=3$, $B=6$, and $C=-53$. Let the image of $P$ be $P'(h, k)$.

We use the image formula: $\frac{h-x_1}{A} = \frac{k-y_1}{B} = -2\frac{Ax_1+By_1+C}{A^2+B^2}$.

First, calculate the term $Ax_1+By_1+C$: $3\left(\frac{17}{3}\right) + 6\left(\frac{28}{3}\right) - 53 = 17 + 2(28) - 53$ $= 17 + 56 - 53 = 73 - 53 = 20$

Next, calculate the term $A^2+B^2$: $3^2 + 6^2 = 9 + 36 = 45$

Now substitute these values into the image formula: $\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = -2\left(\frac{20}{45}\right)$ $\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = -\frac{40}{45}$ Simplify the fraction $-\frac{40}{45}$ by dividing numerator and denominator by 5: $\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = -\frac{8}{9}$

Now, we solve for $h$ and $k$:

• For $h$: $\frac{h - \frac{17}{3}}{3} = -\frac{8}{9}$ $h - \frac{17}{3} = 3 \times \left(-\frac{8}{9}\right)$ $h - \frac{17}{3} = -\frac{24}{9}$ $h - \frac{17}{3} = -\frac{8}{3}$ $h = \frac{17}{3} - \frac{8}{3} = \frac{9}{3} = 3$

• For $k$: $\frac{k - \frac{28}{3}}{6} = -\frac{8}{9}$ $k - \frac{28}{3} = 6 \times \left(-\frac{8}{9}\right)$ $k - \frac{28}{3} = -\frac{48}{9}$ $k - \frac{28}{3} = -\frac{16}{3}$ $k = \frac{28}{3} - \frac{16}{3} = \frac{12}{3} = 4$

Thus, the image of the centroid is $(h, k) = (3, 4)$.

Step 4: Evaluate the Expression $h^2 + k^2 + hk$

Now that we have $h=3$ and $k=4$, we can calculate the required expression: $h^2 + k^2 + hk = (3)^2 + (4)^2 + (3)(4)$ $= 9 + 16 + 12$ $= 25 + 12$ $= 37$

Common Mistakes & Tips

• Be careful with the negative sign in the image formula (the $-2$ term). Forgetting it gives the foot of the perpendicular, not the reflection.
• Double-check your arithmetic, especially when dealing with fractions and solving simultaneous equations.
• When finding the intersection of lines, organize your work to avoid errors in substitution or elimination.

Summary

To solve this problem, we first found the vertices of the triangle by finding the intersection of the given lines. Then, we calculated the centroid of the triangle using the coordinates of its vertices. Finally, we found the image of the centroid in the given line and evaluated the expression $h^2 + k^2 + hk$. The final result is 37.

The final answer is $\boxed{37}$, which corresponds to option (B).