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Given triangle ABC with vertices A(−3, 0), B(0, 6), and C(4, 6). Find the equations of the three medians of triangle ABC

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frika

Answer:

[tex]y=1.2x+3.6\\ \\y=\dfrac{6}{7}x+6\\ \\y=\dfrac{6}{11}x+\dfrac{42}{11}[/tex]

Step-by-step explanation:

Given triangle ABC with vertices A(−3, 0), B(0, 6), and C(4, 6).

First, find midpoints of sides AB, BC and AC:

  • midpoint of AB has coordinates [tex]C_1\left(\dfrac{-3+0}{2},\dfrac{0+6}{2}\right)\rightarrow C_1(-1.5, 3);[/tex]
  • midpoint of BC has coordinates [tex]A_1\left(\dfrac{0+4}{2},\dfrac{6+6}{2}\right)\rightarrow A_1(2,6);[/tex]
  • midpoint of AC has coordinates [tex]B_1\left(\dfrac{-3+4}{2},\dfrac{0+6}{2}\right)\rightarrow B_1(0.5, 3)[/tex]

Now find the eqyations of the medians.

1. Median at vertex A (line [tex]AA_1[/tex]):

[tex]y=\dfrac{6-0}{2-(-3)}(x-(-3))+0\\ \\y=\dfrac{6}{5}(x+3)\\ \\y=1.2x+3.6[/tex]

2. Median at vertex B (line [tex]BB_1[/tex]):

[tex]y=\dfrac{6-3}{4-0.5}(x-0)+6\\ \\y=\dfrac{6}{7}x+6[/tex]

3. Median at vertex C (line [tex]CC_1[/tex]):

[tex]y=\dfrac{6-3}{4-(-1.5)}(x-4)+6\\ \\y=\dfrac{6}{11}(x-4)+6\\ \\y=\dfrac{6}{11}x+\dfrac{42}{11}[/tex]

Answer:

y= -7/6x+6

y= -1/2x+8

x = -3

Step-by-step explanation:

Find the slope of the lines of the triangle. Then you can find the perpendicular  slope of that (the slope of the altitudes) by inversing it and switching the = or - sign. Then you have a point the altitude goes through (one of the vertices) and the slope. Then you can make the equation.

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