Reflecting a point across a line or an axis is a fundamental concept in geometry. Let’s break down the steps to find the reflected point $C’$ of a given point $C$
Reflecting Across the X-Axis
If you have a point $C(x, y)$ and you want to reflect it across the x-axis, the y-coordinate changes its sign while the x-coordinate remains the same. The reflected point $C’$ will be:
$C'(x, -y)$
Example
If $C(3, 4)$, then the reflected point $C’$ across the x-axis is $C'(3, -4)$
Reflecting Across the Y-Axis
If you want to reflect a point $C(x, y)$ across the y-axis, the x-coordinate changes its sign while the y-coordinate remains the same. The reflected point $C’$ will be:
$C'(-x, y)$
Example
If $C(3, 4)$, then the reflected point $C’$ across the y-axis is $C'(-3, 4)$
Reflecting Across the Origin
Reflecting a point $C(x, y)$ across the origin involves changing the signs of both the x and y coordinates. The reflected point $C’$ will be:
$C'(-x, -y)$
Example
If $C(3, 4)$, then the reflected point $C’$ across the origin is $C'(-3, -4)$
Reflecting Across a Line $y = mx + b$
Reflecting a point across a line that isn’t one of the axes is a bit more complex. Let’s consider the general line $y = mx + b$. Here are the steps:
- Find the slope of the perpendicular line: The slope of the line perpendicular to $y = mx + b$ is $-frac{1}{m}$
- Find the equation of the perpendicular line through point $C(x, y)$: This line will have the form $y – y_1 = -frac{1}{m}(x – x_1)$
- Find the intersection point $P$ of the original line and the perpendicular line: Solve the system of equations formed by $y = mx + b$ and $y – y_1 = -frac{1}{m}(x – x_1)$
- Use the midpoint formula: The midpoint between $C$ and $C’$ is $P$. Use the midpoint formula to find $C’$. The midpoint formula is:
$(x_1, y_1) = frac{1}{2}(x + x’, y + y’)$
Solving this will give you the coordinates of $C’$
Example
Let’s reflect the point $C(3, 4)$ across the line $y = x$
- The slope of $y = x$ is $1$, so the slope of the perpendicular line is $-1$
- The equation of the perpendicular line through $(3, 4)$ is $y – 4 = -1(x – 3)$, which simplifies to $y = -x + 7$
- Solve $y = x$ and $y = -x + 7$ simultaneously to find $P$. Solving, we get $x = 3.5$ and $y = 3.5$, so $P(3.5, 3.5)$
- Use the midpoint formula to find $C’$. Since $P$ is the midpoint, $C’$ will be $(4, 3)$
Conclusion
Reflecting a point across various lines or axes involves understanding the geometric transformations that apply to the coordinates. By following these steps, you can accurately find the reflected point $C’$ in different scenarios.