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    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 y-y1/y2/-y1=x-x1/x2-x1

    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
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    For Q2: (2,-5) is (x_1,y_1) and (-7,4) is (x_2,y_2) use m=\frac{y_2-y_1}{x_2-x_1} where m is the gradient to be found. The gradient is therefore \frac9{-9} which = -1.Then use y-y_1=m(x-x_1). So y+5=-1(x-2). Therefore, the equation of the line is y=-x-3. 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 -x-3=0. so  x=-3. P is therefore at (-3,0). 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 m=\frac{y_2-y_1}{x_2-x_1} and find the gradient of the line joining the two. Then sub values into y-y_1=m(x-x_1) 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.
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    (Original post by MostCompetitive)
    For Q2: (2,-5) is (x_1,y_1) and (-7,4) is (x_2,y_2) use m=\frac{y_2-y_1}{x_2-x_1} where m is the gradient to be found. The gradient is therefore \frac9{-9} which = -1.Then use y-y_1=m(x-x_1). So y+5=-1(x-2). Therefore, the equation of the line is y=-x-3. 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 -x-3=0. so  x=-3. P is therefore at (-3,0). 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 m=\frac{y_2-y_1}{x_2-x_1} and find the gradient of the line joining the two. Then sub values into y-y_1=m(x-x_1) 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!
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    (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 m=\frac{y_2-y_1}{x_2-x_1}. Once you've done that, use y-y_1=m(x-x_1), 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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