High School

The following sketch shows triangle ABC with median AM and altitude AP:

A(3,6)
B(1,1)
C(7,-1)

2.1 Determine the equation of AM.

2.2 Determine the length of AM.

2.3 Determine the equation of AP.

2.4 Determine the equation of the perpendicular bisector of AB.

Answer :

To solve the problem regarding triangle ABC where we need to determine various equations and lengths involving median AM and altitude AP, let's break down the question step by step.

Given:

The coordinates of the vertices of the triangle are:

  • A(3, 6)
  • B(1, 1)
  • C(7, -1)

2.1 Determine the equation of AM:

AM is a median, which means it connects vertex A to the midpoint of the side BC. First, we find the midpoint M of line BC.

The coordinates of M are calculated as follows:
[tex]M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)[/tex]
[tex]M = \left( \frac{1 + 7}{2}, \frac{1 - 1}{2} \right) = (4, 0)[/tex]

Now, we use the coordinates of A(3, 6) and M(4, 0) to find the equation of line AM.

The slope of AM is:
[tex]m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 6}{4 - 3} = -6[/tex]

Using point-slope form ([tex]y - y_1 = m(x - x_1)[/tex]):
[tex]y - 6 = -6(x - 3)[/tex]
Simplifying:
[tex]y = -6x + 18 + 6[/tex]
[tex]y = -6x + 24[/tex]

2.2 Determine the length of AM:

To find the length of AM, use the distance formula:
[tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]
[tex]d = \sqrt{(4 - 3)^2 + (0 - 6)^2}[/tex]
[tex]d = \sqrt{1 + 36} = \sqrt{37}[/tex]

2.3 Determine the equation of AP:

AP is an altitude, meaning it is perpendicular to BC. We first calculate the slope of BC:
[tex]m_{BC} = \frac{-1 - 1}{7 - 1} = \frac{-2}{6} = -\frac{1}{3}[/tex]

The slope of AP, being perpendicular to BC, is the negative reciprocal:
[tex]m_{AP} = 3[/tex]

Using point-slope form with A(3, 6):
[tex]y - 6 = 3(x - 3)[/tex]
Simplifying:
[tex]y = 3x - 9 + 6[/tex]
[tex]y = 3x - 3[/tex]

2.4 Determine the equation of the perpendicular bisector of AB:

First, find the midpoint of AB:
[tex]M_{AB} = \left( \frac{3 + 1}{2}, \frac{6 + 1}{2} \right) = (2, 3.5)[/tex]

The slope of AB is:
[tex]m_{AB} = \frac{1 - 6}{1 - 3} = \frac{-5}{-2} = \frac{5}{2}[/tex]

The slope of the perpendicular bisector is the negative reciprocal:
[tex]m = -\frac{2}{5}[/tex]

Using point-slope form with the midpoint:
[tex]y - 3.5 = -\frac{2}{5}(x - 2)[/tex]
Simplifying:
[tex]y = -\frac{2}{5}x + \frac{4}{5} + 3.5[/tex]
[tex]y = -\frac{2}{5}x + \frac{19}{5}[/tex]

This is the set of equations and calculations needed to solve the problem regarding triangle ABC with median AM and altitude AP.