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	$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$	Midpoint
$\overline{AB}$	$A(0,0)$ and $B(6q,6r)$	$\left(\frac{0+6q}{2},\frac{0+6r}{2}\right)$	$L(3q,3r)$
$\overline{BC}$	$B(6q,6r)$ and $C(6p,0)$	$\left(\frac{6q+6p}{2},\frac{6r+0}{2}\right)$	$M(3q+3p,3r)$
$\overline{AC}$	$A(0,0)$ and $C(6p,0)$	$\left(\frac{0+6p}{2},\frac{0+0}{2}\right)$	$N(3p,0)$

Equations of $\overleftrightarrow{AM},$ $\overleftrightarrow{BN},$ and $\overleftrightarrow{CL}$

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

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

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

Line	Slope	Reference Point	$y=mx+b$	$y\text{-}$intercept	Equation
$\overleftrightarrow{AM}$	$\frac{r}{q+p}$	$A(0,0)$	$0=\frac{r}{q+p}\cdot 0+b$	$0$	$y=\frac{r}{q+p}x$
$\overleftrightarrow{BN}$	$\frac{2r}{2q-p}$	$N(3p,0)$	$0=\frac{2r}{2q-p}\cdot 3p+b$	$\text{-}\frac{6pr}{2q-p}$	$y=\frac{2r}{2q-p}x-\frac{6pr}{2q-p}$
$\overleftrightarrow{CL}$	$\frac{r}{q-2p}$	$C(6p,0)$	$0=\frac{r}{q-2p}\cdot 6p+b$	$\text{-}\frac{6pr}{q-2p}$	$y=\frac{r}{q-2p}x-\frac{6pr}{q-2p}$

Coordinates of point $P$

To find the coordinates of point $P,$ we will solve the following system. To do so, we can use the Substitution Method. The $x\text{-}$coordinate of point $P$ is $2q+2p.$ With this, we can find its $y\text{-}$coordinate. Therefore, the coordinates of point $P$ are $(2p+2q,2r).$

Show that point $P$ is on $\overleftrightarrow{CL}$

To verify that point $P$ is the centroid of $△ABC,$ we need to show that $P$ is on $\overleftrightarrow{CL}.$ Since point $P$ satisfies the equation of $\overleftrightarrow{CL},$ 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
$\overline{AM}$	$A(0,0)$ and $M(3q+3p,3r)$	$\sqrt{(3q+3p-0{)}^2+(3r-0{)}^2}$	$\sqrt{(3q+3p{)}^2+(3r{)}^2}$
$\overline{AP}$	$A(0,0)$ and $P(2p+2q,2r)$	$\sqrt{(2p+2q-0{)}^2+(2r-0{)}^2}$	$\sqrt{(2q+2p{)}^2+(2r{)}^2}$

Now, let's check if $\frac{2}{3}AM=AP.$ Since we have a true statement, we can conclude that $\frac{2}{3}AM=AP.$ Proceeding in the same way, we can also show that $\frac{2}{3}BN=BP$ and $\frac{2}{3}CL=CP.$