Exercise 25 Page 417

Let's examine the given diagram.

Using the given diagram, we will prove that the centroid of a triangle is two thirds the distance from each vertex to the midpoint of the opposite side. To develop a proof, we will first find the coordinates of L, M, and N.

Coordinates of L, M and N

To find the coordinates of L, M, and N, we will use the Midpoint Formula.

Side	Endpoints	(x_1+x_2/2,y_1+y_2/2)	Midpoint
AB	A( 0, 0) and B( 6q, 6r)	(0+ 6q/2,0+ 6r/2)	L(3q,3r)
BC	B( 6q, 6r) and C( 6p, 0)	(6q+ 6p/2,6r+ 0/2)	M(3q+3p,3r)
AC	A( 0, 0) and C( 6p, 0)	(0+ 6p/2,0+ 0/2)	N(3p,0)

Equations of AM, BN, and CL

To find the equations of AM, BN, and CL, we will first find their slopes by using the Slope Formula.

Line	Points	y_2-y_1/x_2-x_1	Slope
AM	A( 0, 0) and M( 3q+3p, 3r)	3r- 0/3q+3p- 0	r/q+p
BN	B( 6q, 6r) and N( 3p, 0)	0- 6r/3p- 6q	2r/2q-p
CL	C( 6p, 0) and L( 3q, 3r)	3r- 0/3q- 6p	r/q-2p

Now that we know the slope of each line, we will write their equations in slope-intercept form. y=mx+b In this form, m is the slope and b is the y-intercept. By using one of the points on each line as a reference point, we will compute the y-intercept of each line and write their equations. Let's start!

Line	Slope	Reference Point	y=mx+b	y-intercept	Equation
AM	r/q+p	A( 0, 0)	0=r/q+p* 0+b	0	y=r/q+px
BN	2r/2q-p	N( 3p, 0)	0=2r/2q-p * 3p+b	-6pr/2q-p	y=2r/2q-px-6pr/2q-p
CL	r/q-2p	C( 6p, 0)	0=r/q-2p * 6p+b	- 6pr/q-2p	y=r/q-2px-6pr/q-2p

Coordinates of point P

To find the coordinates of point P, we will solve the following system. y= rq+px & (I) y= 2r2q-px- 6pr2q-p & (II) To do so, we can use the Substitution Method. The x-coordinate of point P is 2q+2p. With this, we can find its y-coordinate. Therefore, the coordinates of point P are (2p+2q,2r).

Show that point P is on CL

To verify that point P is the centroid of △ ABC, we need to show that P is on CL. Since point P satisfies the equation of CL, we can conclude that it is on the line.

Point P is the centroid of △ ABC

To show that point P is two thirds the distance from each vertex to the midpoint of the opposite side, let's begin by finding AM and AP by the Distance Formula.

Side	Endpoints	sqrt((x_2-x_1)^2+(y_2-y_1)^2)	Length
AM	A( 0, 0) and M( 3q+3p, 3r)	sqrt(( 3q+3p- 0)^2+( 3r- 0)^2)	sqrt((3q+3p)^2+(3r)^2)
AP	A( 0, 0) and P( 2p+2q, 2r)	sqrt(( 2p+2q- 0)^2+( 2r- 0)^2)	sqrt((2q+2p)^2+(2r)^2)

Now, let's check if 23AM=AP. Since we have a true statement, we can conclude that 23AM=AP. Proceeding in the same way, we can also show that 23BN=BP and 23CL=CP.