Animated Solution for Mathematics - Straight Lines: The image of the point $(3,5)$ in the line $x-y+1=0$, lies on :

A Mirror Reflection Journey

Imagine you are standing in front of a mirror. You see yourself, but you are not actually there; you are a reflection. In the world of coordinate geometry, this is exactly what we are doing.

We are taking a point $P(3,5)$ and finding its 'twin' on the other side of a mirror line, $x-y+1=0$. This is not just about plugging numbers into a formula; it is about understanding the elegant symmetry of the Cartesian plane.

First, let us look at our mirror. The equation $x-y+1=0$ is our reflecting surface. If we rewrite this as $y=x+1$, we see a line with a slope of $1$ and a y-intercept of $1$.

This is our boundary. Our object point is $P(3,5)$. To find its reflection, we need to find a point $P^{\prime }(x,y)$ such that the line segment $P{P}^{\prime }$ is perpendicular to our mirror line, and the mirror line bisects $P{P}^{\prime }$.

While we could use the slope-midpoint method, we have a more powerful tool in our arsenal: the reflection formula. For a point $(x_1,y_1)$ and a line $ax+by+c=0$, the image $(x,y)$ is found using:

This formula is a masterpiece of efficiency. It handles the perpendicularity and the distance requirements in one go. Let us extract our parameters: $a=1$, $b=-1$, and $c=1$. Our point is $x_1=3$ and $y_1=5$.

Now, let us calculate the position value, $a{x}_1+b{y}_1+c$. Substituting our values, we get $1(3)+(-1)(5)+1=3-5+1=-1$.

Next, we calculate the constant ratio:

With this ratio equal to $1$, the rest is simple algebra. For the x-coordinate: $\frac{x-3}{1}=1\Rightarrow x=4$. For the y-coordinate: $\frac{y-5}{-1}=1\Rightarrow y-5=-1\Rightarrow y=4$.

Our image point $P^{\prime }$ is $(4,4)$.

The problem asks us to find which curve contains this point. We test our point $(4,4)$ against the options.

Let us check Option 1: $(x-2{)}^2+(y-4{)}^2=4$. Substituting $x=4$ and $y=4$, we get $(4-2{)}^2+(4-4{)}^2=2^2+0^2=4$.

The left-hand side equals the right-hand side! We have found our match. The image point lies perfectly on the circle defined by $(x-2{)}^2+(y-4{)}^2=4$.

You have successfully navigated the reflection, proving that with the right tools and a clear geometric vision, even the most daunting problems become simple, elegant truths.