Question

medians of a triangle quick check
triangle abc is drawn on a coordinate plane with vertices a(-2, -3), b(4, 0), and c(-2, 2) and medians indicating the mid - point of each of the line segments ab, bc, and ca. prove that the medians meet at a single point by finding the centroid. (1 point)
(0, 0)
(-1, 1/2)
(0, 1)
(0, -1/4)

Explanation:

Step1: Find mid - point of \(AB\)

The mid - point formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). For \(A(-2,-3)\) and \(B(4,0)\), the mid - point \(D\) of \(AB\) is \((\frac{-2 + 4}{2},\frac{-3+0}{2})=(1,-\frac{3}{2})\).

Step2: Find mid - point of \(BC\)

For \(B(4,0)\) and \(C(-2,2)\), the mid - point \(E\) of \(BC\) is \((\frac{4+( - 2)}{2},\frac{0 + 2}{2})=(1,1)\).

Step3: Find mid - point of \(CA\)

For \(C(-2,2)\) and \(A(-2,-3)\), the mid - point \(F\) of \(CA\) is \((\frac{-2+( - 2)}{2},\frac{2+( - 3)}{2})=(-2,-\frac{1}{2})\).

Step4: Find the equations of the medians

The median from \(C\) to \(AB\) (passing through \(C(-2,2)\) and \(D(1,-\frac{3}{2})\)):
The slope \(m_1=\frac{2+\frac{3}{2}}{-2 - 1}=\frac{\frac{4 + 3}{2}}{-3}=-\frac{7}{6}\). Using the point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(-2,2)\), the equation is \(y-2=-\frac{7}{6}(x + 2)\), or \(y=-\frac{7}{6}x-\frac{7}{3}+2=-\frac{7}{6}x-\frac{1}{3}\).
The median from \(A\) to \(BC\) (passing through \(A(-2,-3)\) and \(E(1,1)\)):
The slope \(m_2=\frac{-3 - 1}{-2 - 1}=\frac{-4}{-3}=\frac{4}{3}\). Using the point - slope form with \((x_1,y_1)=(-2,-3)\), the equation is \(y + 3=\frac{4}{3}(x + 2)\), or \(y=\frac{4}{3}x+\frac{8}{3}-3=\frac{4}{3}x-\frac{1}{3}\).
The median from \(B\) to \(CA\) (passing through \(B(4,0)\) and \(F(-2,-\frac{1}{2})\)):
The slope \(m_3=\frac{0+\frac{1}{2}}{4 + 2}=\frac{1}{12}\). Using the point - slope form with \((x_1,y_1)=(4,0)\), the equation is \(y-0=\frac{1}{12}(x - 4)\), or \(y=\frac{1}{12}x-\frac{1}{3}\).

Step5: Find the intersection point of two medians

Let's find the intersection of the medians from \(A\) and \(C\). Set \(-\frac{7}{6}x-\frac{1}{3}=\frac{4}{3}x-\frac{1}{3}\).
Add \(\frac{7}{6}x\) to both sides: \(0=\frac{4}{3}x+\frac{7}{6}x\), \(\frac{8 + 7}{6}x=0\), \(\frac{15}{6}x=0\), so \(x = 0\).
Substitute \(x = 0\) into \(y=\frac{4}{3}x-\frac{1}{3}\), we get \(y=-\frac{1}{3}\).
We can check if this point lies on the third median. Substitute \(x = 0\) and \(y=-\frac{1}{3}\) into \(y=\frac{1}{12}x-\frac{1}{3}\), \(y=\frac{1}{12}\times0-\frac{1}{3}=-\frac{1}{3}\).

Answer:

\((0,-\frac{1}{3})\)