Question

Recall that a median of a triangle is a segment connecting a vertex of a triangle to the midpoint of the opposite side. Let T be the triangle with vertices$(0,0),(a,0),and(c,d).$In Exercises 70–72, prove the given statements.

The medians of triangle T are concurrent; that is, all three medians intersect at the same point, P.

Short answer

The solution of the equation of medians is the point of the interaction and centroid of the triangle.

So centroid of the triangle is$p(\overline{x},\overline{y})=\left(\frac{a+c}{3},\frac{d}{3}\right).$

Step-by-step solution

Step 1. Given information.

Vertices of given triangle T are$(0,0),(a,0),and(c,d).$

Step 2. Diagram of Triangle

Consider a triangle T of vertices $B(0,0),C(a,0),andA(c,d).$

The midpoint of AB is $Z\left(\frac{c}{2},\frac{d}{2}\right).$

The midpoint ofACis$Y\left(\frac{a+c}{2},\frac{d}{2}\right).$

The midpoint of BCis$X\left(\frac{a}{2},d\right).$

Medians of the triangle are AX, BY,and CZ.

Step 3. Equation of Medians of the triangle 

The equation of median AX is following.

$$\begin{gathered} \frac{y-d}{x-c}=\frac{d-0}{c-\frac{a}{2}} \\ y=\frac{d\left(x-\frac{a}{2}\right)}{c-\frac{a}{2}}\cdots \left(i\right) \end{gathered}$$

The equation of median BY is following.

$$\begin{gathered} \frac{y-\left({\displaystyle \frac{a+c}{2}}\right)}{x-\left({\displaystyle \frac{d}{2}}\right)}=\frac{\left(\frac{a+c}{2}\right)-0}{\left(\frac{d}{2}\right)-0} \\ y=\frac{\left(a+c\right)x}{D}\cdots \left(ii\right) \end{gathered}$$

The equation of median CZ is following.

$$\begin{gathered} \frac{y-0}{x-a}=\frac{\frac{d}{2}-0}{\frac{c}{2}-a} \\ y=\frac{d\left(x-a\right)}{c-2a}\cdots \left(iii\right) \end{gathered}$$

Solution of equations i, ii,and iii is $\left(\frac{a+c}{3},\frac{d}{3}\right).$

The solution of the equation of medians is the point of the interaction and centroid of the triangle.

So centroid of triangle is$p(\overline{x},\overline{y})=\left(\frac{a+c}{3},\frac{d}{3}\right).$