Find the slope of each line and then use the point-slope form to find the equations. To find P, find the intersection point of two lines and check that this point lies on the third line.
See solution.
Practice makes perfect
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.
Finding the Equation of the Lines
We begin by finding the slope of the lines by using the Slope Formula.
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).
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 Substitution Method.
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.
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 Distance Formula, 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.
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 Centroid Theorem.
