Question 957150: A Rectangular parking lot is shown on the site map below.The Entrance and exit points are located at A(-3,-3)B(-5,8) And C(9,1). The Construction crew must place a lamp post so that it will cast a circular light path on the parking lot that will equally light each of the three entrances. Determine the coordinates of where the lamp post should be placed on the grid. Justify.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the 3 exit points form a triangle.
draw a circumscribed circle around this triangle.
the center of the circumscribed circle will be equally distant from all the vertices of the triangle.
your lamp post would need to be placed at the center of that circle.
if the light is strong enough, it will shine equally on all the exits.
how to you find the center of the circumscribed circle?
you need to find the perpendicular bisector of all 3 sides of the triangle.
the intersection of those 3 perpendicular bisectors will be the center of the circle.
the distance between any vertex of the triangle to that intersection will be the center of the circumscribed circle.
the rest is details.
first you need to find the equations of the 3 lines that make up the sides of your triangle.
the line AB will be between (-3,-3) and (-5,8)
the line BC will be between (-5,8) and (9,1)
the line AC will be between (-3,-3) and (9,1)
those equations are:
y = -11/2 * x - 39/2 for line AB.
y = -1/2 * x + 11/2 for line BC.
y = 1/3 * x - 2 for line AC.
next you have to find the midpoints of each of those lines.
those midpoints will be:
(-4,5/2) for midpoint of line AB.
(2,9/2) for midpoint of line BC.
(3,-1) for midpoint of line AC.
next you have to find the slopes for each of the lines perpendicular to lines AB, BC, and AC.
those slopes are the negative reciprocals of the slope of the lines.
those slopes are:
2/11 for the slope of the line perpendicular to AB.
2 for the slope of the line perpendicular to BC.
-3 for the slope of the line perpendicular to AC.
next you have to find the y intercept and the equation of the lines perpendicular to AB, BC, and AC.
those equations are:
y = 2/11 * x + 71/22 for the line perpendicular to AB.
y = 2 * x + 1/2 for the line perpendicular to BC.
y = -3 * x + 8 for the line perpendicular to AC.
when you graph those perpendicular bisectors of the lines AB, BC, AC, you will see that the intersection is at the point (1.5, 3.5)
that's the center of the circumscribed circles.
in order to graph the circle, you need to find the equation of the circle.
this part isn't really necessary but it is done to show you that the intersection of the 3 perpendicular bisectors is the center of the circumscribed circle.
the end of all that is the graph that you see below:
the triangle is the black solid lines.
the perpendicular bisectors and the circumscribed circle are the orange dashed lines.
the reference where i got that you needed the intersection of the perpendicular bisectors is shown below. it has all other kinds of information regarding the center of a triangle as well.
http://jwilson.coe.uga.edu/emt668/EMAT6680.2003.fall/Perry/Assign%204%20Triangle%20Proof/Triangle%20Proof.htm
the formulas that you will need are:
slope intercept formula for a line,.
that formula is y = mx + b.
m is the slope
b is the y-intercept.
the formula to find the slope from 2 points on a line is:
slope = (y2 - y1) / (x2 - x1)
formula to find the slope intercept form of the equation for the line is then found using the point slope formula for a line.
that formula is:
y-y1 = m * (x-x1)
(x1,y1) is any point omn the line and m is the slope that you calculated for the line.
the equation for the midpoint of a line is:
midpoint = ((x1+x1)/2, (y1+y1)/2)
you don't really need to graph the circle so i won't get into the details of how that's done.
suffice it to say that, once you found the intersection point of the perpendicular bisectors of the lines that make up the sides of the triangle, then you are done.
the point turned out to be (1.5,3.5) as shown on the graph.
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