Find the equation of a.Sides b.Median c.Line joining the midpoint of the sides

Question

Given the vertices of a triangle A(0,1) B(6,3) C(3,8). Find the equation of: a.Sides b.Median c.Line joining the midpoint of the sides

Answer 1

Triangle with vertices (0,1), (6,3), and (3,8)

 

To find the equations of the sides we need find the slopes of the lines then 

use the point slope form y-y₁ = m(x-x₁)

 

a.  First find the slopes of the 3 lines:   m = (y₂-y₁)/(x₂-x₁)

    1)  (0,1), (6,3) for AB    m = (3-1)/(6-0) = 2/6 = 1/3

         To use slope point form choose either point... let's use (0,1)

          y-1 = 1/3(x-0)

          y-1 = 1/3(x)

          3y - 3 = x

          AB:  x-3y=-3

 

     2)  (0,1), (3,8) for AC

          m = (8-1)/(3-0) = 7/3

          Again let's use (0,1) with slope 7/3

          y-1 = 7/3(x-0)

          3y-3 = 7x

          AC:  7x-3y = -3

    

     3)  (6,3), (3,8) for BC

          m = (8-3)/(3-6) = -5/2

          Let's use point (6,3) with slope -5/2

          y-3 = -5/2(x-6)

          2y-6 = -5(x-6)

          2y-6 = -5x+30

          BC:  5x+2y = 36  

 

b.  Median..  A triangle has 3 medians.  A median is the line from one vertex to the midpoint of the opposite line.

     The medians intersect in the center of the triangle (known as the "centroid" and each median bisects the

     area of the triangle.

 

    We need to find the midpoint of the 3 sides  AB, AC, and BC

    For the midpoint of two points we use  ((x₁+x_x)/2, (y₁+y₂)/2)

 

    Midpoint of AB (0,1), (6,3)...  ((0+6)/2, (1+3)/2) = (3,2)

    Midpoint of AB is (3,2)

 

    Midpoint of AC (0,1), (3,8)... ((0+3)/2, (1+8)/2) = (3/2, 9/2)

    Midpoint of AC is (3/2, 9/2)

 

    Midpoint of BC (6,3), (3,8) ... ((6+3)/2, (3+8)/2) = (9/2, 11/2)

    Midpoint of BC is (9/2, 11/2)

 

    For the three medians we need to find the equations of the lines joining each vertex to the

    midpoint of the opposite side

 

For median from Vertex A to midpoint of BC we have the line connecting (0,1) and (9/2, 11/2)

m = (11/2 - 1)/(9/2 - 0) = (9/2)/(9/2) = 1

For the point slope formula let's use Point A (0,1) with slope 1

y-1 = 1(x-0)

y-1 = x

x-y=-1

 

For median from B to midpoint of AC we have (6,3), (3/2, 9/2)

m = (9/2 - 3)/(3/2 - 6) = (3/2)/(-9/2) = (3/2)(-2/9) = -1/3

Let's use point B (6,3) and slope -1/3 with point slope formula

y-3 = (-1/3)(x-6)

3y-9 = -1(x-6)

3y-9 = -x+6

x+3y = 15

 

For median of vertex C to midpoint of AB we have (3,8), (3,2)

m = (2-8)/(3-3) = -6/0    slope is undefined. 

This is a vertical line of equation  x = 3

 

Our three medians are:  x-y=-1  ...  x+3y=15  ...   x=3

 

If we need to find the actual centroid of the triangle it is where the medians intersect:

x=3

x-y=-1

x+3y=15

 

Substituting x=3 into the other 2 equations

 

3-y=-1 ...  y=3+1 ...  y = 4

 

3+3y=15 ...  3y=12 ... y=4

 

The centroid of the triangle is the point (3,4)

 

c.  Line joining the midpoint of the sides:

    Our midpoints are   (3,2), (3/2, 9/2), (9/2, 11/2)

 

    We can call these points D, E, F where D = (3,2), E = (3/2, 9/2), F = (9/2, 11/2)

 

    We go through the same process to find the equations of the lines connecting DE, DF, and EF

    Find the slopes of the 3 lines using  (y₂-y₁)/(x₂-x₁)

    Then use the point slope form  y - y₁ = m(x-x₁)

 

I'll leave this to you at this point as we have provided examples above in parts a and b on how to do this.

 

Good luck.  Hope this helps.