Question

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 \( \triangle ABC \) in the line \( 3x + 6y - 53 = 0 \). Then \( h^2 + k^2 + hk \) is equal to:

• A. 47
• B. 36
• C. 40
• D. 37

Hint:

The centroid of a triangle can be found using the intersection of the medians. To find the image of a point across a line, use the reflection formula.

Answer 1

The Correct Option is D

Given two lines:

Equation 1: \( x + 2y - 31 = 0 \)

Equation 2: \( 9x - 2y - 19 = 0 \)

Solving these two equations, we find the points of intersection:

Intersection points are: \( (9,11) \), \( (3,4) \), \( (5,13) \)

The centroid of \( \triangle ABC \) is calculated as:

\[ \text{Centroid} = \left( \frac{17}{3}, \frac{28}{3} \right) \]

Since the image of \( \triangle ABC \) is reflected about the line:

\[ 2x + 6y - 53 = 0 \]

The centroid of the reflected triangle will also be the reflection of the original centroid across this line.

Using the formula for reflection of a point \( (x, y) \) about the line \( ax + by + c = 0 \):

\[ x' = x - \frac{2a(ax + by + c)}{a^2 + b^2}, \quad y' = y - \frac{2b(ax + by + c)}{a^2 + b^2} \]

Substituting \( (x, y) = \left( \frac{17}{3}, \frac{28}{3} \right) \) and \( a = 2 \), \( b = 6 \), \( c = -53 \):

\[ \frac{x - \frac{17}{3}}{2} = \frac{-2\left(2\left(\frac{17}{3}\right) + 6\left(\frac{28}{3}\right) - 53\right)}{2^2 + 6^2} \]

Solving for \( h \) and \( k \):

\[ h = 3, \quad k = 4 \]

Finally, computing:

\[ h^2 + k^2 + hk = (h + k)^2 - hk \]

\[ = 49 - 12 = 37 \]