Answer:
True
Step-by-step explanation:
To answer this question, I will use the following illustrations.
Assume a point (x,y) is reflected across the x-axis.
Using reflection rule, the new point will be: (x,-y)
On a coordinate plane, the x-axis is represented as: (x,0)
So, we will calculate the distance between (x,y) and (x,0) and also calculate the distance between (x,-y) and (x,0) using the following distance formula.
[tex]D= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]
Distance between (x,y) and (x,0)
[tex]D= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]
[tex]D= \sqrt{(x - x)^2 + (0 - y)^2}[/tex]
[tex]D= \sqrt{(0)^2 + (- y)^2}[/tex]
[tex]D= \sqrt{0 + y^2}[/tex]
[tex]D= \sqrt{y^2}[/tex]
[tex]D = y[/tex]
Distance between (x,-y) and (x,0)
[tex]D= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]
[tex]D= \sqrt{(x - x)^2 + (0 - (-y))^2}[/tex]
[tex]D= \sqrt{(0)^2 + (0+y)^2}[/tex]
[tex]D= \sqrt{0 + y^2}[/tex]
[tex]D= \sqrt{y^2}[/tex]
[tex]D = y[/tex]
See that the calculated distance are equal.
Hence, the given statement is true