Animated Solution for Mathematics - Straight Lines: Locus of the image of the point $(2,3)$ in the line $(2x-3y+4)+k(x-2y+3)=0,k\in \mathbb{R}$, is a:

The equation $(2x-3y+4)+k(x-2y+3)=0$ represents a family of lines. In coordinate geometry, the form $L_1+k{L}_2=0$ indicates that every line in the family passes through the intersection of $L_1=0$ and $L_2=0$.

We treat this intersection point as a fixed "hinge," which we will call $C$. All lines in the family rotate around this point $C$.

To locate $C$, we solve the system of linear equations:

From the second equation, we express $x$ as $x=2y-3$. Substituting this into the first equation yields:

Expanding and simplifying, we obtain:

Substituting $y=2$ back into the expression for $x$, we find $x=2(2)-3=1$. Thus, the fixed hinge point is $C(1,2)$.

Consider any generic line $L$ from this family. Since $L$ passes through $C(1,2)$, reflecting point $P(2,3)$ across $L$ to obtain an image $P^{\prime }$ implies that $L$ is the perpendicular bisector of the segment $P{P}^{\prime }$.

A fundamental property of the perpendicular bisector is that any point on it is equidistant from the segment's endpoints. Because $C$ lies on the mirror line $L$, it must be equidistant from $P$ and $P^{\prime }$.

This leads to the invariant geometric condition:

Since $C(1,2)$ and $P(2,3)$ are both fixed, the distance $CP$ is a constant. We calculate this distance as follows:

Because $CP=\sqrt{2}$, it follows that $C{P}^{\prime }=\sqrt{2}$ for every line in the family. As the line $L$ rotates around $C$, the image $P^{\prime }$ maintains a constant distance of $\sqrt{2}$ from $C$.

A point moving at a constant distance from a fixed center traces a circle. Therefore, the locus of $P^{\prime }$ is a circle centered at $C(1,2)$ with a radius of $\sqrt{2}$.