Question

Locus of the image of the point $(2, 3)$ in the line $\left(2x - 3y + 4\right) + k\left(x - 2y + 3\right) = 0, k \in R,$ is a

• A. Straight line parallel to x-axis
• B. Straight line parallel to y-axis
• C. Circle of radius $\sqrt{2}$
• D. Circle of radius $\sqrt{3}$

Answer 1

The Correct Option is C

To find the locus of the image of the point \((2, 3)\) in the given line, let's follow these steps:

1. Consider the equation of the line \(\left(2x - 3y + 4\right) + k\left(x - 2y + 3\right) = 0\), where \(k \in \mathbb{R}\).
2. Simplify the equation:
   • Expanding the equation: \(2x - 3y + 4 + kx - 2ky + 3k = 0\).
   • Combine like terms: \((2 + k)x + (-3 - 2k)y + (4 + 3k) = 0\).
   • This represents a family of lines depending on \(k\).
3. The perpendicular distance from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is given by: \(\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
4. Calculate the image of the point \((2, 3)\) about the variable line:
   • Finding the perpendicular line from \((2, 3)\) to \((2+k)x + (-3-2k)y + (4+3k) = 0\) and its reflection forms the image.
   • The perpendicular line from \((2, 3)\) on this line will have direction ratios \(2+k\) and \(-3-2k\), meaning the line perpendicular will have a gradient \(\frac{3 + 2k}{2 + k}\).
5. The equation and solution of this image construction are complex, so let's analyze the geometric property:
   • Since \(k\) is a parameter, the family of lines is changing. The reflection of the point across each line generates a locus.
   • Notice that the coefficients \((2+k)\) and \((-3-2k)\) change such that any pair \((x', y')\) for reflection will keep a constant distance relation.
   • By observing the symmetry and simplified distance, the locus of reflection defines a fixed radius.
6. This geometric property ultimately forms a circle in the plane.
7. Thus, the solution provides: The locus is a circle with a radius of \(\sqrt{2}\).

Hence, the correct answer is: Circle of radius \(\sqrt{2}\).