Question

Will medal & fan :-)
Triangle ABC has vertices A(0,0), B(2,2), C(5,-1). Find the coordinates of L, the midpoint of AC , and M, and the midpoint of BC . Verify that LM||AB and LM = 1/2 AB .
Please help i am beyond confused /_\

Answer 1

This is just an application of the Midpoint Formula and a bit of expansion on it with the second part. First, use the midpoint formula to find L and M:

\[Midpoint (x,y)=(\frac{ x _{1}+x _{2} }{ 2 },\frac{ y _{1}+y _{2} }{ 2 })\]
For the second part, it helps to draw a picture of the triangle the three points create, and you will see that LM is parallel to AB, to verify this you need yo find the slope of LM and AB (I can give you the slope formula if you need it), and then you need to find the distance using the formula for distance between two points for LM and AB and you should get that LM=1/2AB. Start with the midpoints L and M and if you need more help I will work with you on the rest of it.

Answer 2

@Alphabet_Sam is the midpoint for AC (2.5, .5) & BC (3.5,.5) ? Also I don't understand the whole lm and ab point , except that lm is half of ab

Answer 3

The midpoint for AC, which we called L should have a negative y-coordinate since it's in the second quadrant. The midpoint for BC, which we called M is (3.5,.5) so that's right. Here's a picture of what we're looking at to help you understand what I wrote in my previous post:

Answer 4

Sorry, the midpoint L is in Quadrant 4 not 2. That was my mistake in what I wrote above.

Answer 5

@Alphabet_Sam so AB = (2,0) and LM =(1,0) ?

Answer 6

So those are lengths of lines, and when you put something in (x,y) form it's a point. You're looking for a length, which will be a regular number like 2 and 1, and the way you find it is with the distance between two points formula:
\[d=\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{1})^{2}}\]

Answer 7

My first sentence was worded poorly. What you found were coordinates of points, but AB and LM represent the lengths of lines.

Answer 8

@Alphabet_Sam so I'd put those coordinates into the distance formula

Answer 9

Yup, the coordinates of A and B for AB and the coordinates of L and M for LM.

Answer 10

@Alphabet_Sam for LM i got 37 is that correct ?

Answer 11

So for LM your two points should be L(2.5,-.5) and M(3.5,.5) which gives you:
\[d=\sqrt{(2.5-3.5)^{2}+(.5-(-.5))^{2}}\]\[d=\sqrt{(-1)^{2}+(1)^{2}}\]\[d=\sqrt{2}\]You try AB now and use this as an example.