# Tuesday, Mar 21

Hi All –

In class today, I passed back a number of tests which I graded over the weekend and were on the progress reports yesterday. We also studied properties of reflections and perpendicular bisectors. Students took notes on how to find the equation of a perpendicular bisector line between any two points. This relates to reflections (the reflection line is a perpendicular bisector between any point and its reflection) and the perpendicular bisector can also be used to find the center of rotation.

Question: How would you find the equation of the perpendicular bisector line for the points A(4, 2) and B(2, 7)?

Answer: First find the slope between these two points $\dpi{300}\inline \dfrac{7 - 2}{2- 4} = \dfrac{5}{-2} = -\dfrac{5}{2}$, now note that the slope of a line perpendicular must be the negative reciprocal slope, so our perpendicular bisector will have slope $\dpi{300}\inline -\left(-\dfrac{2}{5}\right) = \dfrac{2}{5}$.

Now, we need to find the midpoint of A and B – so we find the average of their x and y- coordinates: $\dpi{300}\inline \left(\dfrac{4+2}{2}, \dfrac{2+7}{2}\right) = (3, 4.5)$

Ok, so we know the slope of the perpendicular bisector and one of the points on the line, we can use point-slope form to write the equation of the line in the form y = m(x – x1) + y1

$\dpi{300}\inline y = \dfrac{2}{5}(x - 3) + 4.5$

If we want this equation in slope-intercept (y = mx + b) form, we just need to simplify the equation above.

$\dpi{300}\inline y = \dfrac{2}{5}(x) + \dfrac{2}{5}(-3) + 4.5$ (distribute the 2/5)

$\dpi{300}\inline y = \dfrac{2}{5}x - \dfrac{6}{5} + \dfrac{9}{2}$ (simplify 2/5 times -3)

$\dpi{300}\inline y = \dfrac{2}{5}x - \dfrac{12}{10} + \dfrac{45}{10}$ (common denominators for like terms)

So, finally, in slope-intercept form, the perpendicular bisector between points A and B is

$\dpi{300}\inline y = \dfrac{2}{5}x + \dfrac{33}{10}$

or

$\dpi{300}\inline y = 0.4x + 3.3$

Today's assignment: is to complete the "Reflections are neat!" packet for tomorrow.