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# working out co-ordinates? Watch

1. Sorry if this sounds ridiculously easy, but i've just stumbled across this exercise in chapter 5 of my c1 textbook and i'm completely baffled on how to work these sets of questions out, I know you have to use the equation

Basically 2,3,4 are finding the "co ordinates"

5,6 are equation of line joining points

7,8 are passing through points

9,10, there aren't any examples in the book!

I've just had a collapse of concentration and I would appreciate any help on what method I should use to work out 2-4 and 5-6 mainly, thank you in advanced
2. For Q2: (2,-5) is and (-7,4) is use where m is the gradient to be found. The gradient is therefore which = -1.Then use . So . Therefore, the equation of the line is . You want to find where it meets the x-axis, so put y equal to zero (because every point on the x axis y=0). So . so . P is therefore at . Do the same for 3 and 4 but, instead of putting y equal to zero, put x equal to zero (because at every point on the y axis x=0).
For 5, put x and y equal to zero for each of the equations to find the co-ordinates of A and B. Then use and find the gradient of the line joining the two. Then sub values into to find the equation of the line. It is the same for 6 except you have to put everything one one side and make it equal to zero once you find the equation of the line.
3. (Original post by MostCompetitive)
For Q2: (2,-5) is and (-7,4) is use where m is the gradient to be found. The gradient is therefore which = -1.Then use . So . Therefore, the equation of the line is . You want to find where it meets the x-axis, so put y equal to zero (because every point on the x axis y=0). So . so . P is therefore at . Do the same for 3 and 4 but, instead of putting y equal to zero, put x equal to zero (because at every point on the y axis x=0).
For 5, put x and y equal to zero for each of the equations to find the co-ordinates of A and B. Then use and find the gradient of the line joining the two. Then sub values into to find the equation of the line. It is the same for 6 except you have to put everything one one side and make it equal to zero once you find the equation of the line.
Thank you for clearing this up, I was able to do most of them up to question 6! your method is a good alternative to the other way because this involves an additional step but I think I find this easier

if you know the method for question 7 as well that would be excellent, but if you don't then don't worry you've helped me alot already!
4. (Original post by d_aan)
Thank you for clearing this up, I was able to do most of them up to question 6! your method is a good alternative to the other way because this involves an additional step but I think I find this easier

if you know the method for question 7 as well that would be excellent, but if you don't then don't worry you've helped me alot already!
For question 7, you solve the equations simultaneously to find the x co-ordinates of S. You then substitute the value of x into either equation and find the y-co-ordinate. Then you find the gradient of the line you want by using . Once you've done that, use , where m is the gradient you've just found, to find the equation of the line. It doesn't matter which of the co-ordinates you use (you can use s or t)
EDIT: If and when you need some more assistance (now/in the future), send me a private message. I'll be happy to help

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Updated: December 10, 2010
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