Question

Prove using coordinate geometry: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

Answer 1

Given: Line l is the perpendicular bisector of CD .
Prove: Point R(a, b) is equidistant from points C and D.

Answer 2

so you want to prove RC =RD

do you have all the co-ordinates ? of R,C,D ?

Answer 3

hint : C is origin

Answer 4

I just need to prove R(a, b) is equidistant from points C and D.
I don't know the coordinates

Answer 5

R(a,b)
means (a,b) are the co-ordinates of R

Answer 6

Oh, wow..Should have thought about it..And C would be (0,0) Right?

Answer 7

correct

now you know the distance formula ?

Answer 8

Umm..no..

Answer 9

Distance between points (x1,y1) and (x2,y2) is
\(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

Answer 10

Never mind..didn't even see that.lol..ok..

Answer 11

How would i know what coordinates to plug in?

Answer 12

lets say we want to find RD
so we need distance between R (a,b) and D(2a,0)

so x1,y1 = a,b
x2,y2 = 2a,0

makes sense ?

Answer 13

Oh..Ok. Yes. That definitely makes sense:)

Answer 14

try and find RD and RC
see whether they come out equal.

Answer 15

Ok. Give about like 5 sec:)

Answer 16

take 50 :)

Answer 17

Hhaha..Ok..I'm trying..

Answer 18

Ok. I don't know if I got this right though...
\[d=\sqrt{(ab-2a,0)^2+(ab-2a,0)^2}\]

Answer 19

x1 is not ab

x1,y1 = a,b
which means
x1 = a, y1 = b

Answer 20

x2 =2a
y2 =0

Answer 21

ooo..my bad..ugu..I don't know..I give up...

Answer 22

Wait. kidding let me try again:)

Answer 23

try once more ?
since now you already have
x1,x2,y1,y2

Answer 24

Yes. I will:)

Answer 25

Ok. Here's what I got..
\[d=\sqrt{(a-2a)^2+(b-0)^2}\]

Answer 26

correct, continue :)

Answer 27

Yay!!! You're going to have to show me..Because...idk.

Answer 28

a-2a = ... ?

Answer 29

-2? or like -2a?

Answer 30

1-2 = ... /

Answer 31

1

Answer 32

2-1 = .. ?

Answer 33

It's 1. Isn't it?

Answer 34

if 2-1 is = 1
then how come
1-2 also =1
?

Answer 35

Lol...ok..you found my mistake:)
2-1 = 1.
1-2 = -1.
I though we didn't have to apply the integer rules.

Answer 36

ok

so a -2a = a(1-2) = a(-1) = -a

makes sense ?

Answer 37

Yes. It does.

Answer 38

what is
\(\Large (-a)^2 = ...?\)

Answer 39

It would be a.

Answer 40

sure ?
\((-a)^2 = (-a)\times (-a) = ... ?\)

Answer 41

Ok. It would be -a then..

Answer 42

two negatives become positive

\(-1 \times -1 = +1\)

so, \(-a \times -a = +a\times a =+a^2\)
see if you get this

Answer 43

Oh yea. Ok. I get it..

Answer 44

and b-0 = b
so,
\(RD = \sqrt {a^2+b^2}\)

right ?

now similarly, can you find RC ?

Answer 45

Yes. Ok..Let me try now. Give me about 5 sec:)

Answer 46

i think you'll need 50

Answer 47

Hahha:) yea. I think i do too.

Answer 48

Ok. I have
\[d=\sqrt{(a-2a)^2+(0,0)}\]

Answer 49

oh, isn't that CD ?

we need RC

R (a,b)
C (0,0)

Answer 50

x1 = a
y1 =b
x2= 0
y2= 0

Answer 51

I thought that was RC..Oh well. hang on..

Answer 52

\[d=\sqrt{(a-0)^2+(b-0)^2}\]

Answer 53

correct
continue

whats a-0
whats b-0

Answer 54

a
b

Answer 55

plug those in

do you get RC = RD ?

Answer 56

Hahhahahaha!!!!!:) Yes!!!!!
\[d=\sqrt{(a^2+b^2)+(a^2+b^2)}\]

Answer 57

RC =RD
means
Point R(a, b) is equidistant from points C and D.

Answer 58

\(\Large RC = \sqrt{a^2+b^2}\)
\(\Large RD = \sqrt{a^2+b^2}\)

Answer 59

So.that's all?

Answer 60

yes,
you proved what you wanted

you proved RC=RD
which means
Point R(a, b) is equidistant from points C and D.

Answer 61

Yay!!!! Now I'm out. I need a break..lol!!:)

Answer 62

me too :)

Answer 63

Yea. I agree!!:) Definitely!!!!
\(\huge\cal\color{lime}{Thank~you~soooooooo~much!!!!!!}\)

Answer 64

\[ \begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#E6E600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#E6E600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array} \]