Okay, so I have a few problems to do, but I have no clue how to solve them. It is geometry proofs using the distance formula I will post the pictures here
there are the first 3 problems. If anyone is patient enough to stay and help me with them until I get it would be greatly appreciated
do you know the formula to find slope given 2 points?
y2-y1 ------= slope? x2-x1
looks right. how would you go about finding the slope of AC?
A(0,0) C(a,a) a-0 ----= ? a-0
1
you can simplify it because a-0=a and a/a=1
okay, I get it that part.
try bd
B(a,0) D(0,a) a-0 a ---- = --= -1? 0-a -a
correct. what do you know about the slopes of perpendicular lines?
You have to take the reciprocal of the original slope. Like 1/3 would be -3/1 or just -3
exactly. so we have slopes 1 and -1 of AC/BD respectively
|dw:1425853107279:dw| what do es that mean
Line AC is perpendicular to line ___ would it be BD
yep
wow, okay that was easy. Now it's the other ones that scare me haha
the 2nd one looks bleh :/ let's start with P use the fact that the midpoint for \((x_1,y_1) \text {and}(x_2,y_2)\) is \((\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})\)
okay, I know how to solve for midpoint but p has a two in it and I'm not sure where that is coming from
what do you mean?
P= 0+2a\2 Where is the 2a part from on the diagram?
P is the midpoint of A and B, right? so our x coordinate is half (you can think of it as the average) of the sum of the 2 x coordinates 0+2a
the 0 is the x coord of A the 2a is the x coord of B
ooohhh, okay! so they're meeting in the middle of P I was thinking it was using the points of P. Makes so much more sense now.
do you want to fill in the rest of the stuff for P?
or does anything else need explaining
I need to kind of be walked through it. This thing online is so confusing so when someone is explaining it me it makes more sense.
i'll try writing it out then show it to you
cool, cool that works
|dw:1425854475056:dw|
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