# Finding equation of a line

• Oct 17th 2006, 11:20 AM
Universe
Finding equation of a line
Hello,
I can't understand 2 questions from my math homework can you please explain how I am supposed to do that?
All you need to do is just find the equations.

1. Find the equations of the medians of the triangle with vertex coordinates at J(2,-2) K(4,-1) and L(-2,-5).

2. Find the equation of the perpendicular bisector of a chord of a circle, given that the end points of the chord are C(-2,0) and D(4,-4).

Thanks a lot!
• Oct 17th 2006, 11:43 AM
Jameson
a) Ok. So there are three midpoints. One on JK, one on JL, and another on KL.

You said you just need a line through the midpoint. There infinitely many that cross through a single point, but we can assume that another point on the line is at the origin for simplicity.

So the midpoint of JL is the sum of the coordinates divided by 2. So x = (2-2)/2 = 0 and y = (-2-5)/2 = -7/2. So now you have the midpoint (0, -7/2). Now you can use any other point to make a line. I'd use the origin to make it simple. So you have two points, (0,0) and (0, -7/2). That's all you need for the line.

Repeat the process for the other two midpoints.
• Oct 17th 2006, 01:33 PM
Universe
Thanks!
but I need an equation for these lines. I mean y=mx+b.
I just can't understand how it supposed to be done =\
any suggestions?
• Oct 17th 2006, 01:38 PM
Jameson
y=mx+b is only one form of a line.

The point-slope form is how the line is derived.

Slope = m.

m= (y-y_0) / (x-x_0) This is the definition of the slope.

So rewriting, y-y_0=m(x-x_0).

You have two points on the line, so use that to find the slope. The plug in x_0, y_0, and m into the equation above.