Coordinate Geometry - Moving Points: The Magic of Reflections-advanced

Find the coordinates of the image of point $P(3, -4)$ after reflection in the line $x = -2$.

Given point $P(x, y) = (3, -4)$ and line of reflection $x = a$ where $a = -2$. The formula for reflection in $x = a$ is $(2a - x, y)$. Substituting the values: $x' = 2(-2) - 3 = -4 - 3 = -7$ $y' = -4$ So, the image $P'$ is $(-7, -4)$.

In a reflection across a vertical line $x = a$, the vertical distance from the axis remains zero (the $y$-coordinate is unchanged), and the horizontal distance from the line $x = a$ is preserved on the opposite side.

A point $A(2, 3)$ is reflected in the $x$-axis to $A'$. $A'$ is then reflected in the line $y = 2$ to $A''$. Find the coordinates of $A''$.

Step 1: Reflection of $A(2, 3)$ in the $x$-axis. Using $M_x(x, y) = (x, -y)$, we get $A'(2, -3)$. Step 2: Reflection of $A'(2, -3)$ in the line $y = 2$. Using $M_{y=b}(x, y) = (x, 2b - y)$, where $b = 2$: $x'' = 2$ $y'' = 2(2) - (-3) = 4 + 3 = 7$ Therefore, $A'' = (2, 7)$.

Successive reflections are applied one after the other. The first transformation maps the original point to an intermediate image, which then serves as the 'object' for the second transformation.

Find the point $P$ on the $x$-axis such that the distance $AP + PB$ is minimum, where $A(1, 2)$ and $B(5, 4)$.

To minimize $AP + PB$ where $P$ is on the $x$-axis (line $y=0$):

1. Reflect point $A(1, 2)$ in the $x$-axis to get $A'(1, -2)$.
2. The shortest distance is the straight line connecting $A'(1, -2)$ and $B(5, 4)$.
3. Find the equation of line $A'B$: Slope $m = \frac{4 - (-2)}{5 - 1} = \frac{6}{4} = \frac{3}{2}$. Equation: $y - 4 = \frac{3}{2}(x - 5) \Rightarrow 2y - 8 = 3x - 15 \Rightarrow 3x - 2y = 7$.
4. Point $P$ lies on the $x$-axis, so $y = 0$. $3x - 2(0) = 7 \Rightarrow 3x = 7 \Rightarrow x = \frac{7}{3}$. Point $P$ is $(\frac{7}{3}, 0)$.

This is an application of Heron's Principle. Reflecting one point makes the path $A'PB$ a single straight line, which is the shortest distance between two points.

1. Reflection of $B(5, 2)$ in $y = -1$: Using $M_{y=b}(x, y) = (x, 2b - y)$ with $b = -1$: $B' = (5, 2(-1) - 2) = (5, -4)$
2. Reflection of $B'(5, -4)$ in the $y$-axis: Using $M_y(x, y) = (-x, y)$: $B'' = (-5, -4)$

We first use the formula for reflection across a horizontal line to find $B'$, then apply the standard $y$-axis reflection rule to the intermediate result.

1. Reflection of $C(-2, 5)$ in the origin $O$: $C' = (-(-2), -5) = (2, -5)$
2. Reflection of $C'(2, -5)$ in the line $x = 3$: Using $M_{x=a}(x, y) = (2a - x, y)$ with $a = 3$: $C'' = (2(3) - 2, -5) = (6 - 2, -5) = (4, -5)$

Origin reflection changes signs of both coordinates. Reflection in $x=a$ modifies the $x$-coordinate based on the distance from the vertical line while keeping the $y$-coordinate constant.