Posted by **chris** on Wednesday, February 18, 2009 at 9:52pm.

the verticies of triangle R(0,12), S(-6,6), and T(4,8). write in slope-intercept form the equations of the lines that contain the segments described.

it says, find the medians of triangle RST.

find the altitudes of triangle RST.

find the perpendicular bisectors of triangle RST.

- math -
**Reiny**, Wednesday, February 18, 2009 at 10:26pm
This is a lengthy question, and it would be unreasonable for you to expect somebody to do it for you.

I will give you a few hints and steps that you can follow

Median: - a line from a vertex to the midpoint of the opposite side

I will do the median from R to side TS

midpoint of TS = M((-6+4)/2,(8+6)/2) = M(-1,7)

slope of RM = (7-12)/(-1-0) = 5

so equation for RM is y = 5x + b

plug in (-1,7)

7 = -5 + b

b = 12 , so equation is y = 5x + 12

(we could have used 12 directly for b, since (0,12) was the y-inercept)

Altitude: - a line from a vertex to the opposite side, or the opposite side extended, meeting that side at right angles.

let's do the altitude from R to ST

slope of ST = 2/5

so the slope of the altitude is -5/2

equation is y = (-5/2)x + b

(0,12) is on it

12 = (-5/2)(0) + b

b = 12

equation , y = (-5/2)x + 12

(posting what I have so far, to be continued)

- math -
**Reiny**, Wednesday, February 18, 2009 at 10:30pm
Perpendicular bisector - a line from the midpoint of a line segment, at right angles to it.

so to find the perpendicular bisector of RS

1. find the midpoint of RS

2. find the slope of RS

3. the slope of the new line will be the negative reciprocal of the slope of RS

4. now you have the slope of that line and a point on it, find its equation.

You should have 3 equations for each of your questions, or 9 in total

I gave you 2 of them, plus the steps for the last type.

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