We will follow the hint and split the proof into three parts: finding the equation of the lines containing the medians, finding the coordinates of P, and finding the distances.
We begin by finding the slope of the lines by using the .
m = y_2-y_1/x_2-x_1
Let's find the slope of the line passing through A(0,0) and R(3b+3a,3c).
Let's substitute (x_1,y_1) = A(0,0) and m_1= cb+a.
y-y_1 = m(x-x_1)
y- 0 = c/b+a(x- 0)
y = c/b+ax
Consequently, the equation of AR is y= cb+ax. Using the same process, we can find the equation of BS and CQ. We summarize the computations in the table below.
| Points
|
m=y_2-y_1/x_2-x_1
|
y-y_1 = m(x-x_1)
|
Equation
|
| B(6b,6c) and S(3a,0)
|
m_2 &= 0 - 6c/3a - 6b
&= 2c/2b-a
|
y-0 = 2c/2b-a(x-3a)
|
BS: y = 2c/2b-a(x-3a)
|
| C(6a,0) and Q(3b,3c)
|
m_3 &= 3c - 0/3b - 6a
&= c/b-2a
|
y-0 = c/b-2a(x-6a)
|
CQ: y = c/b-2a(x-6a)
|
Finding the Coordinates of P
To find P, let's find the intersection point between AR and BS by solving the system of equations below.
y= cb+ax & (I) y= 2c2b-a(x-3a) & (II)
Let's solve this system by applying the .
y= cb+ax & (I) y= 2c2b-a(x-3a) & (II)
y= cb+ax cb+ax= 2c2b-a(x-3a)
y= cb+ax (2b-a)cx=2c(b+a)(x-3a)
y= cb+ax (2b-a)x=2(b+a)(x-3a)
y= cb+ax 2bx-ax=2bx-6ab+2ax-6a^2
y= cb+ax -3ax=-6ab-6a^2
y= cb+ax x=2b+2a
To find the y-coordinate of P we substitute x=2b+2a into Equation (I).
y=c/b+ax x=2b+2a
y=c/b+a( 2b+2a) x=2b+2a
â–¼
(I): Simplify right-hand side
y=2c x=2b+2a
Consequently, the coordinates of P are (2b+2a,2c). Let's verify that CQ passes through P by substituting its coordinates into the corresponding equation.
y = c/b-2a(x-6a)
2c ? = c/b-2a( 2b+2a-6a)
2c ? = c/b-2a(2b-4a)
2c ? = 2c/b-2a(b-2a)
2c = 2c
As we can see, we got a true statement which implies that CQ passes through P. Consequently, the three medians intersect at point P.
Finding the Distances
By using the , we will find the length of each median and also the distance from each vertex to P.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
Let's begin by finding the length of AR.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
AR = sqrt(( 3b+3a- 0)^2+( 3c- 0)^2)
AR = sqrt((3b+3a)^2+(3c)^2)
AR = sqrt((3(b+a))^2+(3c)^2)
AR = sqrt(3^2(b+a)^2+3^2c^2)
AR = sqrt(3^2((b+a)^2+c^2))
AR = 3sqrt((b+a)^2+c^2)
Next, let's find the length of AP.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
AP = sqrt(( 2b+2a- 0)^2+( 2c- 0)^2)
AP = sqrt((2b+2a)^2+(2c)^2)
AP = sqrt((2(b+a))^2+(2c)^2)
AP = sqrt(2^2(b+a)^2+2^2c^2)
AP = sqrt(2^2((b+a)^2+c^2))
AP = 2sqrt((b+a)^2+c^2)
Let's divide the latter equation by 2.
AP/2 = 2sqrt((b+a)^2+c^2)/2 = sqrt((b+a)^2+c^2)
Next, we can substitute this equation into AR=3sqrt((b+a)^2+c^2).
AR=3sqrt((b+a)^2+c^2)
⇒
AR &= 3AP/2
⇒
AP &= 2/3AR
The latter equation tells us that P is two thirds of the distance from A to R. Next, let's find the length of the two remaining medians and the distance from B and C to P. We summarize all the computations in the table below.
| Points
|
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
|
Length
|
| B(6b,6c) and S(3a,0)
|
BS = sqrt((3a-6b)^2+(0-6c)^2)
|
BS = 3sqrt((a-2b)^2 + 4c^2)
|
| B(6b,6c) and P(2b+2a,2c)
|
BP = sqrt((2b+2a-6b)^2+(2c-6c)^2)
|
BP = 2sqrt((a-2b)^2 + 4c^2)
|
| C(6a,0) and Q(3b,3c)
|
CQ = sqrt((3b-6a)^2+(3c-0)^2)
|
CQ = 3sqrt((b-2a)^2 + c^2)
|
| C(6a,0) and P(2b+2a,2c)
|
CP = sqrt((2b+2a-6a)^2+(2c-0)^2)
|
CP = 2sqrt((b-2a)^2 + c^2)
|
Next, we divide the BP and CP by 2 and then substitute the resulting equations into BS and CQ respectively.
BP/2 = sqrt((a-2b)^2 + 4c^2) & ⇒ BP = 2/3BS [0.3cm]
CP/2 = sqrt((b-2a)^2 + c^2) & ⇒ CP = 2/3CQ
In consequence, we can affirm that P is two thirds of the distance from each vertex to the midpoint of the opposite side, which proves the .
