Question Statement

Find the equations of the medians and the point of concurrency of the triangle $DEF$ when $D(-6,-4)$, $E(6,-4)$ and $F(-2,4)$. What is the name of the point of concurrency?

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Background and Explanation

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. The three medians intersect at a single point called the point of concurrency. To solve this problem, you need the midpoint formula and the two-point form of a line equation.

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Solution

Consider the triangle $DEF$ with vertices $D(-6,-4)$, $E(6,-4)$, and $F(-2,4)$:

Step 1: Find the Midpoints of the Sides

Let $M_1$, $M_2$, and $M_3$ be the midpoints of sides $\overline{DE}$, $\overline{EF}$, and $\overline{FD}$ respectively. Using the midpoint formula $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$:

$\begin{array}{rl}{M}_1& =\left(\frac{-6+6}{2},\frac{-4+(-4)}{2}\right)=(0,-4)\\ M_2& =\left(\frac{6+(-2)}{2},\frac{-4+4}{2}\right)=(2,0)\\ M_3& =\left(\frac{-6+(-2)}{2},\frac{-4+4}{2}\right)=(-4,0)\end{array}$

Step 2: Equation of Median $\overline{D{M}_2}$

The median from $D(-6,-4)$ to $M_2(2,0)$. Using the two-point form $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$:

$\begin{array}{rl}\frac{x-(-6)}{2-(-6)}& =\frac{y-(-4)}{0-(-4)}\\ \frac{x+6}{8}& =\frac{y+4}{4}\\ \frac{x+6}{2}& =\frac{y+4}{1}\text{(dividing both sides by 4)}\\ x+6& =2y+8\\ x-2y-2& =0\end{array}$

Step 3: Equation of Median $\overline{E{M}_3}$

The median from $E(6,-4)$ to $M_3(-4,0)$:

$\begin{array}{rl}\frac{x-6}{-4-6}& =\frac{y-(-4)}{0-(-4)}\\ \frac{x-6}{-10}& =\frac{y+4}{4}\\ \frac{x-6}{-5}& =\frac{y+4}{2}\text{(dividing both sides by 2)}\\ 2(x-6)& =-5(y+4)\\ 2x-12& =-5y-20\\ 2x+5y+8& =0\end{array}$

Step 4: Equation of Median $\overline{F{M}_1}$

The median from $F(-2,4)$ to $M_1(0,-4)$:

$\begin{array}{rl}\frac{x-(-2)}{0-(-2)}& =\frac{y-4}{-4-4}\\ \frac{x+2}{2}& =\frac{y-4}{-8}\\ \frac{x+2}{1}& =\frac{y-4}{-4}\text{(dividing both sides by 2)}\\ -4(x+2)& =y-4\\ -4x-8& =y-4\\ 4x+y+4& =0\end{array}$

Step 5: Finding the Point of Concurrency

To find where the medians intersect, solve any two of the equations (i), (ii), and (iii). Let's solve equations (i) and (ii):

Multiply equation (i) by 2: $2x-4y-4=0$

Subtract equation (ii) from this result: $\begin{array}{rl}(2x-4y-4)-(2x+5y+8)& =0\\ -9y-12& =0\\ 9y& =-12\\ y& =-\frac{4}{3}\end{array}$

Substitute $y=-\frac{4}{3}$ into equation (i): $\begin{array}{rl}x-2\left(-\frac{4}{3}\right)-2& =0\\ x+\frac{8}{3}-2& =0\\ x+\frac{8}{3}-\frac{6}{3}& =0\\ x& =-\frac{2}{3}\end{array}$

Thus, the point of concurrency is $\boxed{\left(-\frac{2}{3},-\frac{4}{3}\right)}$.

Step 6: Verification and Identification

We can verify this result using the centroid formula, which states that the point of concurrency of medians has coordinates $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$:

$\left(\frac{-6+6+(-2)}{3},\frac{-4+(-4)+4}{3}\right)=\left(\frac{-2}{3},\frac{-4}{3}\right)$

This confirms our solution. The point of concurrency of the medians of a triangle is called the centroid of the triangle.

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Key Formulas or Methods Used

• Midpoint Formula: $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
• Two-Point Form of a Line: $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$
• Centroid Formula: $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$
• Solving Simultaneous Linear Equations: Substitution or elimination method

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Summary of Steps

1. Calculate midpoints $M_1$, $M_2$, $M_3$ of the three sides $\overline{DE}$, $\overline{EF}$, and $\overline{FD}$ using the midpoint formula.
2. Find equations of the three medians $\overline{D{M}_2}$, $\overline{E{M}_3}$, and $\overline{F{M}_1}$ using the two-point form of a line.
3. Solve any two median equations simultaneously using elimination or substitution to find the intersection point.
4. Verify the result using the centroid formula $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$.
5. Identify the point of concurrency as the centroid of the triangle.