# why aren't reflected lines perpendicular, meaning why are their slopes negative not negative reciprocals?

*print*Print*list*Cite

### 1 Answer

The change in the slope when you reflect a line depends on the line of reflection:

(1) If you reflect across the y-axis you replace x with (-x) resulting in **opposite slopes**.

Ex If you reflect the line y=2x+3 across the y-axis you will get the line y=2(-x)+3 or y=-2x+3

Note what happens in the slope formula -- changing all x values to their opposites ==> ` `The slope of the original line is `m=(y_2-y_1)/(x_2-x_1)` while the slope of the reflected line is `m_"*"=(y_2-y_1)/((-x_2)-(-x_1))=-(y_2-y_1)/(x_2-x_1)`

(2) If you reflect across the x-axis, you replace y with (-y) also resulting in **opposite slopes**.

Ex If you reflect y=2x+3 across the x-axis you get -y=2x+3 or y=-2x-3

(3) If you reflect a line across the line y=x you get **reciprocal slopes**. Ex If you reflect y=2x+3 across the line y=x you get y=1/2x-3/2

Reflecting across the line y=x is equivalent to exchanging all x and y values. The slope for the reflected line is `m_"*"=(x_2-x_1)/(y_2-y_1)=1/m`

All of these assume that the given line is not parallel to or perpendicular to the reflecting line. Other cases exist for different lines of reflection. It is possible to find a line of reflection for a given line that produces a reflection perpendicular to the given line. For example reflecting the line y=x+3 across the y-axis results in a perpendicular reflection:

**Sources:**