Question

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Find the coordinates of C in terms of b.

x=
y=

Answer 1

i know that b is 2 but how do i find C?

Answer 2

to find C put y= x^2 ...you'll have quadratic in x ..then solve for x to get C

Answer 3

how do you know b is 2? and if you know the equation of the line, finding C should come easily... i suppose. don't you think?

Answer 4

how do i solve for x to get C? ghazi

Answer 5

you have equation of line...by two points and that line passes through C ...you can find C by using Y=x^2 ...first write equation of line

Answer 6

im not sure if b is 2 now sasogeek

Answer 7

you have (1,1) and (0,b) ...your line will be \[(y-1)=\frac{ (b-1) }{ 0-1 }*(x-1)\]

Answer 8

now put y=x^2 in the above equation and solve for s

Answer 9

*x

Answer 10

oh ok

Answer 11

you'll have x in terms of b and hence y

Answer 12

to find x, you should know what b is :)

Answer 13

the equation of the line is (b-1)/(0-1)*(x-1)+1 and the other equation is x^2 so they should be equal at C

Answer 14

so to solve for C you set x^2=(b-1)/(0-1)*(x-1)+1

Answer 15

right?

Answer 16

yes

Answer 17

do i solve for B?

Answer 18

you don't have to have values in terms of figures though, find C in terms of b so the coordinates of C will have b in it.

Answer 19

you don't have to solve for b

Answer 20

solve for x in terms of b

Answer 21

how do i solve for x in terms of b?

Answer 22

i have xs in both sides

Answer 23

wait a sec

Answer 24

\[x^2+(b-1)x-b=0\] now solve for x

Answer 25

i get \[(1-b)(x-1)-x^2+1=0\]

Answer 26

not what you got

Answer 27

and x=-b

Answer 28

I got the same thing.

Answer 29

and for y?

Answer 30

so x=-b and y=b^2

Answer 31

literally (-b)^2, but the negative doesn't matter in the square

Answer 32

yeah i got that -b^2

Answer 33

coordinate of point C are (x,y) ..now x=b , y= x^2=b^2 (b,b^2)

Answer 34

but yeah your right its b^2

Answer 35

x=-b

Answer 36

well according to the figure x should be = -b

Answer 37

yeah

Answer 38

thanks guys

Answer 39

:)