To reflect a point across the line y = 2, we simply take the y-coordinate of the point and subtract it from 2?
The Correct Answer and Explanation is :
To reflect a point across the line ( y = 2 ), you indeed need to consider the y-coordinate of the point, but the process involves more than just subtracting it from 2. Here’s the correct approach:
- Identify the Point: Let’s say you have a point ( P(x, y) ) that you want to reflect across the line ( y = 2 ).
- Find the Distance to the Line: The line ( y = 2 ) is horizontal, so the distance from the point ( P ) to the line is given by ( d = y – 2 ) (if ( y ) is above the line) or ( d = 2 – y ) (if ( y ) is below the line).
- Calculate the Reflected Point: The reflected point ( P’ ) will be located the same distance from the line ( y = 2 ) but on the opposite side. Thus, the y-coordinate of the reflected point ( P’ ) can be calculated as follows:
- If ( y ) is greater than 2 (above the line):
[ P'(x, y’) = \left(x, 2 – (y – 2)\right) = \left(x, 4 – y\right) ] - If ( y ) is less than 2 (below the line):
[ P'(x, y’) = \left(x, 2 + (2 – y)\right) = \left(x, 4 – y\right) ]
In both cases, the formula simplifies to the same result for the reflected point:
[ P'(x, y’) = (x, 4 – y) ]
Example: If the point is ( P(3, 5) ):
- The distance from ( P ) to the line ( y = 2 ) is ( 5 – 2 = 3 ).
- To reflect it, calculate the new y-coordinate:
[ P'(3, y’) = (3, 4 – 5) = (3, -1) ]
So the reflected point across the line ( y = 2 ) is ( P'(3, -1) ).
This process can be visualized geometrically: the original point and its reflection are equidistant from the line ( y = 2 ), ensuring a symmetric relationship with respect to the line of reflection.