This is one of the more tedious (some might say difficult, but I believe it is just more work to be done in order to come up with the final solution) question in the high school geometry curriculum. You are given the midpoints of a triangle, and you have to find the coordinates of the actual triangle. For our example, we are given the midpoints of the triangle as A (-4, 1) B (-2, 2) C (-5, 3) and we will need to find the vertices. For most students, I believe having a visual (mental) image of the actual triangle that would encompass these midpoints will help to clarify the problem just a little better.
In order to "sketch" the actual triangle, you will need to find 3 lines that are parallel to the lines created by the triangle using the midpoints. So for line segment AB, there will be a parallel line which will include the point C (from now on called Line1). There will also be a line segment parallel to BC that will include point A (Line2), and the same goes for a parallel line to AC that will include point B (Line3).
This type of problem will ask you to recall information you have learned back in Algebra 1. First you need to be able to write an equation of a line, given 2 points. Then you will need to solve the intersection of 2 lines, which means that it will provide you with the coordinates of the intersection. This intersection between 2 lines is one of the vertex of the triangle. And that also means that you will need to do repeat this process for the other 2 vertices (using the combination of the other 2 lines).
So let's find the equation of Line1. First we need to find out the slope of Line1. Since Line1 is parallel to AB, that means that if we find the slope of AB, we have then found the slope Line1. If we know the slope of the Line1, we can use point C to find the equation of Line1.
A (-4, 1) B (-2, 2)
Slope of AB = (2 - 1)/(-2 + 4) = 1/2
Using slope of 1/2, and point C (-5, 3), we can find the equation of Line1.
y = mx + b
3 = 1/2(-5) + b
b = 11/2
Line1: y = x/2 + 11/2
B (-2, 2) C (-5, 3)
Slope of BC = (3 - 2)/(-5 + 2) = 1/-3 = -1/3
Using slope of 1/2, and point A (-4, 1), we can find the equation of Line2.
y = mx + b
1 = -1/3(-4) + b
b = -1/3
Line2: y = -x/3 - 1/3
A (-4, 1) C (-5, 3)
Slope of AC = (3 - 1)/(-5 + 4) = 2/-1 = -2
Using slope of -2, and point B (-2, 2), we can find the equation of Line3.
y = mx + b
2 = -2(-2) + b
b = -2
Line3: y = -2x - 2
Now that we the lines of the 3 lines, we will need to solve the 3 linear equations 3 different times.
Let's solve Line1 intersecting Line2:
Line1: y = x/2 + 11/2
Line2: y = -x/3 - 1/3
x/2 + 11/2 = -x/3 - 1/3
x/2 + x/3 = -1/3 - 11/2
5x/6 = (-2 - 33)/6
5x = -35
x = -7
Solve y using either Line1 or Line2:
y = 1/2(-7) + 11/2
y = -7/2 + 11/2 = 4/2 = 2
Vertex1 = (-7, 2)
Let's solve Line2 intersecting Line3:
Line2: y = -x/3 - 1/3
Line3: y = -2x - 2
-2x - 2 = -x/3 - 1/3
-2x + x/3 = -1/3 + 2
(-6x + x)/3 = (-1 + 6)/3
-5x = 5
x = -1
Solve for y using Line2 or Line3:
y = -2x - 2 = -2(-1) - 2 = 2 - 2 = 0
Vertex2 = (-1, 0)
Let's solve Line1 intersecting Line3:
Line1: y = x/2 + 11/2
Line3: y = -2x - 2
x/2 + 11/2 = -2x - 2
x/2 + 2x = -2 - 11/2
(x + 4x)/2 = (-4 - 11)/2
5x = -15
x = -3
Substitute for y using either Line1 or Line3:
y = -2x - 2 = -2(-3) - 2 = 4
Vertex3 = (-3, 4)
If you have any question regarding this type of problems, please feel free to reach out to me or any of the instructors in my center.
Michael Huang
Center Director
Mathnasium of Glen Rock/Ridgewood
T: 201-444-8020
E: glenrock@mathnasium.com
www.mathnasium.com/glenrock
