Q. 70 - Calculus

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.

Step-by-Step Solution

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 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 $B(0, 0), C(a, 0), and A(c, d).$

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

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

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

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