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

Two lines are provided:

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

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

The intersection points of these lines are determined to be: \( (9,11) \), \( (3,4) \), and \( (5,13) \).

The centroid of \( \triangle ABC \) is found to be: \( \left( \frac{17}{3}, \frac{28}{3} \right) \).

The triangle \( \triangle ABC \) is reflected across the line:

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

The centroid of the reflected triangle is the reflection of the original centroid across this line.

The formula for reflecting a point \( (x, y) \) across the line \( ax + by + c = 0 \) is given by:

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

With \( (x, y) = \left( \frac{17}{3}, \frac{28}{3} \right) \) and line parameters \( a = 2 \), \( b = 6 \), \( c = -53 \), the reflection is calculated as:

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

The coordinates \( h \) and \( k \) of the reflected centroid are found to be: \( h = 3 \), \( k = 4 \).

The final computation is:

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

\[ = 49 - 12 = 37 \]