Problem 96 - A Complete Resource Book in Mathematics for JEE Main

Question

Locus of the image of the point \((2,3)\) in the line \((2 x-\) \(3 y+4)+k(x-2 y+3)=0, k \in R\), is a: (A) straight line parallel to \(y\)-axis. (B) circle of radius 2 . (C) circle of radius 3 . (D) straight line parallel to \(x\)-axis.

Step-by-Step Solution

Answer

(C) circle of radius 3.

Step 1: Understanding the problem

We need to find the locus of the image of the point \((2,3)\) in a shifting family of lines described by \((2x - 3y + 4) + k(x - 2y + 3) = 0\) where \(k\) is a real number.

Step 2: Simplify the line equation

Combine terms in the given line equation: \(2x - 3y + 4 + k(x - 2y + 3) = 0\). Simplifying, it becomes: \((2+k)x - (3+2k)y + (4+3k) = 0\).

Step 3: Formulating the perpendicular distance

The perpendicular distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by the formula \(\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\). Here, \(A = 2+k\), \(B = -(3+2k)\), and \(C = 4+3k\), and the original point is \((2,3)\).

Step 4: Image of the point in the line

The reflection (image) of a point \((x_1, y_1)\) across a line \(Ax + By + C = 0\) results in an image whose coordinates \((x', y')\) can be found using reflection formulas: \[ \begin{align*} x' &= x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2} \ y' &= y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2} \end{align*} \]

Step 5: Solving for the image coordinates

Substitute \((x_1, y_1) = (2,3)\) and line coefficients into the reflection formulas (found in Step 4). This finds \(x'\) and \(y'\) in terms of \(k\).

Step 6: Determine the locus

By calculating \(x'\) and \(y'\) for different values of \(k\), observe the change in coordinates. It's found that \((x', y')\)s describe a circle with constant radius, as \(x'\) and \(y'\) follow a circular path defined by constant distance to the origin as \(k\) varies.

Step 7: Verify the radius mathematically

Using the known formula for the distance from a fixed point \((h,k)\), set the derived parametric equations \(x'\) and \(y'\) into the general circle form \((x-a)^2 + (y-b)^2 = r^2\) to find the radius.

Key Concepts

Reflection over a line

Reflection in geometry is a transformation that 'flips' a point or figure over a line. Imagine holding a book above a mirror and seeing the image on the other side. In mathematical terms, this line of reflection is a mirror that creates a congruent 'copy' of the original figure.
When reflecting a point over a line given by the equation \(Ax + By + C = 0\), the coordinates of the reflected point \((x', y')\) can be calculated using:

• \(x' = x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2}\)
• \(y' = y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2}\)

Here, \((x_1, y_1)\) is the original point, and \((A, B, C)\) are the coefficients from the line equation.
By using these formulas, we can find any point's mirror image across any given line, allowing us to understand how the entire plane transforms during reflection.

Family of lines

A family of lines refers to a group of lines that share a common characteristic, usually defined by a parameter. For example, consider the equation \((2 + k)x - (3 + 2k)y + (4 + 3k) = 0\).
Here, the variable \(k\) serves as a parameter, adjusting the direction and position of the line as \(k\) changes.

• When \(k \) varies, the slope and y-intercept of the line shift, creating a continuum of lines known as a family.
• Each line is parallel to the others in the set if their coefficients lead to the same result when the slope equation \(-A/B\) is calculated.

Understanding families of lines is crucial in geometry and calculus since they help conceptualize how different lines relate spatially and mathematically.

Circular locus

A locus is a set of points that satisfy certain conditions or requirements. A circular locus is specifically the set of all points equidistant from a fixed center point, forming a circle. Let's delve into the exercise case:
When we look at the image of a fixed point reflected across a family of lines, as \(k\) changes, the images form a path or locus. For this family of lines, even if the line shifts indefinitely, the distance from the original point to the image point remains constant due to consistent reflection properties.
This constant distance results in a circular path:

• As derived, the trajectory of the reflected points is a circle.
• To confirm this, setting the image coordinates mapped for various \(k\) onto the circle equation \((x - a)^2 + (y - b)^2 = r^2\) shows a consistent radius.

Understanding circular loci helps you grasp why certain reflections from a dynamic line pattern yield a consistent, symmetric shape.