2.7 - Using Coordinates to Solve Problems

Rebecca is designing a parking lot. A tall light will illuminate the three entrances, at A, B and C. Rebecca needs to position the lamp so that it illuminates each entrance equally.

A(-8, 14)

B(-4, 8)

C(18, 10)

How can Rebecca determine the location of the lamp?

Example #1 cont’d

A(-8, 14)

B(-4, 8)

C(18, 10)

We are looking for the circumcenter since this is the point that is an equal distance from each vertex.We need to draw 3 perpendicular bisectors and find where they intersect – this is the circumcenter

The midpoint of AB is:

Each purple line has a slope which is the negative reciprocal of the slope of the red line it originates from

mAB = Therefore, the slope of the perpendicular bisector is An equation is y1 = x + b

12

3

Example #1 cont’dAn equation is y1 = x + bWe can use the point (-6, 11) to find b:11 = + b

11 = -4 + bb = 15

Therefore, the equation of the perpendicular bisector of AB is y1 = x + 15

Example #1 cont’d

A(-8, 14)

B(-4, 8)

C(18, 10)

12

3

The midpoint of BC is:

mBC = Therefore, the slope of the perpendicular bisector is -11An equation is y2 = -11x + b

We can use the point (7, 9) to find b:y2 = -11x + b9 = -11(7) + b9 = -77 + bb = 86Therefore the equation of the perpendicular bisector of BC is y2 = -11x + 86

Example #1 cont’dTwo lines already give us the point of intersection so we don’t have to find ACOur 2 lines are:

y1 = x + 15y2 = -11x + 86

Point of Intersection is where y1 = y2

x + 15 = -11x + 8611x = x = 71 x = 6.09y = -11(6.09) + 86 = 19.01

Therefore, if the lamp is placed at (6.09, 19.01), it will be about the same distance from each entrance.

Example #1 Part 2

Power line

Cable

Lamp (6, 19)

(12, 10)

(0, 4)

The shortest distance from the lamp to the power line is a perpendicular distance.

m =

Therefore, the equation of the power line is y = x + 4

The cable is perpendicular to the power line, so the slope of the cable is -2.

An equation for the perpendicular line is y = -2x + b

A point on the line is (6, 19):19 = -2(6) + b19 = -12 + bb = 31y = -2x + 31

Example #1 – Part 2 cont’d

Power line

Cable

Lamp (6, 19)

(0, 4)

We need to find where the Power Line and Cable Line intersect:x + 4 = -2x + 312.5x = 27x = 10.8y = -2(10.8) + 31y = 9.4

Therefore the cable should be connected to the power line at (10.8, 9.4).

Length of cable == = = 10.73 Therefore about 10.8m of cable will be needed to connect the lamp to the power line.