Q. 72 - 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.

Prove that the centroid of triangle T is two-thirds of the way from each vertex to the opposite side.

Step-by-Step Solution

Answer

the centroid of triangle T is two-thirds of the way from each vertex to the opposite side.

$$\begin{gathered} \frac{AP}{AX}=\frac{\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}}{\sqrt{{\left(c-\frac{a}{2}\right)}^2+d^2}\begin{array}{}\\ \\ \\ \end{array}} \\ \frac{AP}{AX}=\frac{\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}}{{\displaystyle \frac{3}{2}\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}}} \\ \frac{AP}{AX}=\frac{2}{3} \end{gathered}$$

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.

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

Step 3. Distance between centroid and vertex

Determine the distance between vertex A and centroid P.

$$\begin{gathered} AP=\sqrt{{\left(c-\frac{a+c}{3}\right)}^2+{\left(d-\frac{d}{3}\right)}^2} \\ AP=\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2} \end{gathered}$$

Distance between vertex A and midpoint X.

$$\begin{gathered} AX=\sqrt{{\left(c-\frac{a}{2}\right)}^2+(d-0{)}^2} \\ AX=\sqrt{{\left(c-\frac{a}{2}\right)}^2+d^2}\begin{array}{}\\ \\ \\ \end{array} \end{gathered}$$

The ratio of AP to AX.

$$\begin{gathered} \frac{AP}{AX}=\frac{\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}}{\sqrt{{\left(c-\frac{a}{2}\right)}^2+d^2}\begin{array}{}\\ \\ \\ \end{array}} \\ \frac{AP}{AX}=\frac{\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}}{{\displaystyle \frac{3}{2}}\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}} \\ \frac{AP}{AX}=\frac{2}{3} \end{gathered}$$

So centroid of triangle T is two-thirds of the way from each vertex to the opposite side.