Question

Use the two-point form to find the equation of each median of AABC. (Recall that a median of a triangle is the line drawn from any vertex to the midpoint of the opposite side.)A(4,3), B(0,7), C(-1,2)Find the equation of each altitude of AABC. (Recall that an altitude of a triangle is the perpendicular drawn from any vertex to the opposite side.) A(4,3), B(0,7), C(-1,2)

Answer 1

Answer:

• 3y+x=13

,

• y+3x=7

,

• y-x=3

Explanation:

Given points A(4,3), B(0,7), C(-1,2):

First, we find the coordinates of the midpoints of AB, AC and BC.

$\begin{gathered} \text{Midpoint of AB}=\mleft(\frac{4+0}{2},\frac{3+7}{2}\mright)=(2,5) \\ \text{Midpoint of AC}=\mleft(\frac{4-1}{2},\frac{3+2}{2}\mright)=(1.5,2.5) \\ \text{Midpoint of BC}=\mleft(\frac{0-1}{2},\frac{7+2}{2}\mright)=(-0.5,4.5) \end{gathered}$

Part 1

The equation of the median from A to BC.

Using points: A(4,3) and (-0.5,4.5) in the two-point form, we have:

$\begin{gathered} \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} \\ \frac{y-3}{x-4}=\frac{4.5-3}{-0.5-4} \\ \frac{y-3}{x-4}=\frac{1.5}{-4.5} \\ \frac{y-3}{x-4}=-\frac{1}{3} \\ 3(y-3)=-1(x-4) \\ 3y-9=-x+4 \\ 3y+x=4+9 \\ 3y+x=13 \end{gathered}$

Part 2

The equation of the median from B to AC.

Using points: B(0,7) and (1.5,2.5) in the two-point form, we have:

$\begin{gathered} \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} \\ \frac{y-7}{x-0}=\frac{2.5-7}{1.5-0} \\ \frac{y-7}{x}=\frac{-4.5}{1.5} \\ \frac{y-7}{x}=-3 \\ y-7=-3x \\ y+3x=7 \end{gathered}$

Part 3

The equation of the median from C to AB.

Using points: C(-1,2) and (2,5) in the two-point form, we have:

$\begin{gathered} \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} \\ \frac{y-2}{x+1}=\frac{5-2}{2+1} \\ \frac{y-2}{x+1}=\frac{3}{3} \\ \frac{y-2}{x+1}=1 \\ y-2=x+1 \\ y-x=1+2 \\ y-x=3 \end{gathered}$

The equations of the medians are:

• 3y+x=13

,

• y+3x=7

,

• y-x=3