You need to find the midpoints of each side of triangle such that:

`D ((x_B+x_C)/2 , (y_B+y_C)/2)=gtD ((30+0)/2 , (0+12)/2)=gt D (15, 6)`

`E((x_A+x_C)/2 , (y_A+y_C)/2)=gtE ((28+0)/2 , (40+12)/2)=gt E(14, 26) `

`F((x_A+x_B)/2 , (y_A+y_B)/2)=gtF ((28+30)/2 , (40+0)/2)=gt F(29, 20) `

Since you know the midpoints and the vertices, you need to use the form of equation that passes through two points such that:

`y - y_D = ((y_A - y_D)/(x_A - x_D))*(x - x_D)`

`y - 6 = (40-6)/(28-15)(x - 15)`

`y - 6 = 34/13(x - 15)`

`y = (34/13)x - 432/13`

`y - y_E = ((y_B - y_E)/(x_B - x_E))*(x - x_E)`

`y - 26 = (0 - 26)/(30 - 14)(x - 14)`

`y - 26 = -(26/16)(x - 14)`

`y - 26 = -(13/8)(x - 14)`

`y = -(13/8)x + 182/8 + 26`

`y = -(13/8)x + 390/8`

`y = -(13/8)x + 195/4`

`y - y_F = ((y_C - y_F)/(x_C - x_F))*(x - x_F)`

`y - 20 = (12 - 20)/(0 - 29)(x - 29)`

`y = (8/29)x- 232/29 + 20`

`y = (8/29)x + 348/29`

**Hence, evaluating the equations of medians of triangle ABC under given conditions yields `y = (34/13)x - 432/13 ` (AD median) ; `y = -(13/8)x + 195/4` (BE median) ; `y = (8/29)x + 348/29 ` (CF median).**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now