*reflecting line,*to make the transformation. In coordinate geometry, the reflecting line is indicated by a lowercase

*l*.

This figure illustrates an important property of reflecting lines: If you form segment *RR'* by connecting pre-image point *R* with its image point *R'* (or *P* with *P'* or *Q* with *Q'*), the reflecting line, *l,* is the perpendicular bisector of segment *RR'*.

A reflecting line is a perpendicular bisector. When a figure is reflected, the reflecting line is the perpendicular bisector of all segments that connect pre-image points to their corresponding image points.

Here's a problem that uses this idea: In the following figure, triangle*J'K'L'*is the reflection of triangle

*JKL*over a reflecting line. Find the equation of the reflecting line using points

*J*and

*J'*. Then confirm that this reflecting line sends

*K*to

*K'*and

*L*to

*L'*.

The reflecting line is the perpendicular bisector of segments connecting pre-image points to their image points. Because the perpendicular bisector of a segment goes through the segment's midpoint, the first thing you need to do to find the equation of the reflecting line is to find the midpoint of line segment *JJ'*:

Next, you need the slope of line segment *JJ'*:

Now you can finish the first part of the problem by plugging the slope of 2 and the point (5, 6) into the point-slope form for the equation of a line:

That's the equation of the reflecting line, in slope-intercept form.

To confirm that this reflecting line sends *K* to *K'* and *L* to *L'*, you have to show that this line is the perpendicular bisector of line segments *KK'* and *LL'*. To do that, you must show that the midpoints of line segments *KK'* and *LL'* lie on the line and that the slopes of line segments *KK'* and *LL'* are both –1/2 (the opposite reciprocal of the slope of the reflecting line, *y* = 2*x* – 4). First, here's the midpoint of line segment *KK'*:

Plug these coordinates into the equation *y* = 2*x* – 4 to see whether they work. Because 12 = 2 (8) – 4, the midpoint of line segment *KK'* lies on the reflecting line. Now get the slope of line segment *KK'*:

This is the desired slope, so everything's copasetic for *K* and *K'*. Now compute the midpoint of line segment *LL'*:

Check that these coordinates work when you plug them into the equation of the reflecting line, *y* = 2*x* – 4. Because 10 = 2(7) – 4, the midpoint of line segment *LL'* is on the line. Finally, find the slope of line segment *LL'*:

This checks. You're done.