Question

In any triangle, a segment that joins a vertex to the midpoint of the opposite side is called a median. Although it passes through the midpoint of the side, it may or may not be perpendicular to that side, so a median is not the same as a perpendicular bisector. Let = (− 2, 0), = (6, 0), and = (4, 6).
a. Find the equation for the median drawn from A to BC
b. Find the equation for the median drawn from B to AC
c. Find the equation for the median drawn from C to AB
d. Show that the medians are concurrent. This point is called the centroid of the triangle.

Answer 1

Answer:

  a. 3x -7y = -6

  b. 3x +5y = 18

  c. 3x -y = 6

  d. concurrent point (8/3, 2) satisfies all equations

Step-by-step explanation:

Given triangle ABC with vertices (-2, 0), (6, 0), (4, 6), you want the equations of the median lines, and you want to show they are concurrent.

Midpoints

The midpoint of each segment is the average of the two endpoints of that segment:

  D (on AB) = (A +B)/2 = (-2+6, 0+0)/2 = (2, 0)

  E (on BC) = (B +C)/2 = (6+4, 0+6)/2 = (5, 3)

  F (on CA) = (C +A)/2 = (4-2, 6+0)/2 = (1, 3)

Equations

The equation for a line through points (x1, y1) and (x2, y2) can be written as ...

  $\Delta y\cdot x-\Delta x\cdot y=-y_1\cdot\Delta x+x_1\cdot\Delta y\\\\(y_2-y_1)x-(x_2-x_1)y=-y_1(x_2-x_1)+x_1(y_2-y_1)$

The first section of the third attachment shows the calculations of the differences ∆x and ∆y for each median. The second section of the third attachment shows the corresponding calculation of the right-side constant (c) in the equation of the line.

a. A to BC

From the third attachment, we have ∆x = 7, ∆y = 3, and c = -6.

The equation for the median from A is 3x -7y = -6.

b. B to AC

From the third attachment, we have ∆x =  -5, ∆y = 3, and c = 18.

The equation for the median from B is 3x +5y = 18.

c. C to AB

From the third attachment, we have ∆x = -2, ∆y = -6, and c = -12. This gives us the equation -6x +2y = -12. We can divide this by -2 to put it in standard form.

The equation for the median from C is 3x -y = 6.

d. Centroid

Any pair of these equations can be solved for the coordinates of the centroid. For example, subtracting the third equation from the first, we have ...

  (3x -7y) -(3x -y) = (-6)-(6)

  -6y = -12

  y = 2

From the last equation, ...

  3x -2 = 6 . . . . . . . . substitute 2 for y

  3x = 8 . . . . . . . add 2

  x = 8/3

The coordinates of the centroid are (8/3, 2).

The last section of the third attachment shows these coordinates satisfy all three equations of the medians. (For the purpose, the constant is subtracted, so the equations are in general form. The result of 0 is the result we want.) For example, ...

  3x +5y -18 = 0
  3(8/3) +5(2) -18 = 0
  8 +10 -18 = 0 . . . . . . . true

showing the centroid point satisfies the second equation. (We already know the point satisfies the other two equations, because we used those to find the centroid.)

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Additional comment

The math is pretty tedious, using three points to find 3 midpoints, then those 6 points to find equations for three lines. That is why we have used calculator vector functions where possible. Even so, keeping the x, y, ∆x, and ∆y values straight, so they are all used properly in the calculations, takes a bit of mental gymnastics. We have tried to show enough detail above so you can see how we did it.

The fist attachment shows the triangle, its medians, their equations, and the centroid. Using a geometry program to find midpoints and write the equations for the lines was a big help. (Centroid x-coordinate 2.67 is the decimal approximation of 8/3.)