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    MATHS HELP PLEASE!
    Please can anyone help me with my homework! The question is:

    Three points, A, B, and C have co-ordinates (1,3), (3,5) and (-1, y). Find the values of y when:

    i) AB = AC
    ii) AC = BC
    iii) AB is perpendicular to BC
    iiii) A, B and C are collinear


    Now I have attempted 'i' and I got the answer 1 which is correct, or it can be 5 (but I don't know how you would get that ) however I moved on to 'ii' and got 1 again but my textbook says it should be '7'. Please can someone help me understand! I am also struggling to understand the last two.
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    I) so work out AB using pythag, then work out AC algebraically (same method - so the difference in y values will be algebraic e.g 3-y) and by using the value of AB as the value of AC in that equation (ab=ac) you get a quadratic with y. Solve to get 2 answers.


    ii) work out the length of AC and BC as an algebraic expression (with y) using pythag and equate (so again, you will know the value for the difference in the x values but the difference in y values will be in terms of y) . Solve to find y.

    iii) by using the coordinates, find the gradients of the lines. Again, this will be algebraic as you don't know y. When you get the algebraic expressions for the gradients, use the fact that if they are perpendicular then they must multiply to make -1. Then solve for y.

    iv) collinear means they are all on the same straight line. Therefore the gradient for AC and AB, and in fact CB, must all be equal.

    hope this makes sense?
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    (Original post by hhattiecc)
    I) so work out AB using pythag, then work out AC algebraically (same method - so the difference in y values will be algebraic e.g 3-y) and by using the value of AB as the value of AC in that equation (ab=ac) you get a quadratic with y. Solve to get 2 answers.


    ii) work out the length of AC and BC as an algebraic expression (with y) using pythag and equate (so again, you will know the value for the difference in the x values but the difference in y values will be in terms of y) . Solve to find y.

    iii) by using the coordinates, find the gradients of the lines. Again, this will be algebraic as you don't know y. When you get the algebraic expressions for the gradients, use the fact that if they are perpendicular then they must multiply to make -1. Then solve for y.

    iv) collinear means they are all on the same straight line. Therefore the gradient for AC and AB, and in fact CB, must all be equal.

    hope this makes sense?
    I'm sorry this has confused me a lot, aha
    If you could write the formula to support the instructions as I am not sure how you would go about using Pythag in the answer. I managed to get the answer 1 for question 'i' but not 5 so I'm guessing the only way to gain both answers is by Pythag? If you don't mind add in the sums for each step that would be useful
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    (Original post by basketballgirl11)
    I'm sorry this has confused me a lot, aha
    If you could write the formula to support the instructions as I am not sure how you would go about using Pythag in the answer. I managed to get the answer 1 for question 'i' but not 5 so I'm guessing the only way to gain both answers is by Pythag? If you don't mind add in the sums for each step that would be useful
    Sorry for confusing you! Hopefully this will clear things up. So plot A and B on a graph. If you draw a line from A to B (i.e AB) then make a right angled triangle, you can work out the length of AB by using Pythagoras’ theorem with AB as the hypotenuse. So the difference in x values is one side length of your triangle, and the same with the y values (if you’re struggling to see this then look at your diagram). So difference in x values is 3-1=2. Difference in y values is 5-3=2. Using pythag: 2^2 + 2^2 =AB^2 = 8.
    You can do the same with AC but this time the difference in y values will be in terms of y. Difference in x=1- -1=2. Difference in y=3-y. Using pythag: 2^2 +(3-y)^2 = AC^2. AC=AB therefore AC^2 also equals 8. Hence you have an equation of 4 + (3-y)^2 = 8. This is just an equation you can solve for y, but you cansee that the answers of 1 and 5 are the solutions by subbing in those values.

    Take exactly the same approach with AC = BC. So (difference in x values)^2 +(difference in y values)^2 = (distance between two points)^2… I think this is a formula you need to know in A level. Should be in your course book I think.

    With the gradients… Change in y/change in x. So for finding the gradient of the line AB you should be able to do, and as for BC use the same equation: change in y =5-y, change in x=3- -1=4. So gradient for that is (5-y)/4. Perpendicular gradients have a product of -1 (again, just something you should know from your studies) so (gradient of AB) x (gradient ofBC) = -1. This will give you an algebraic equation in terms of y, which shouldbe pretty straight forward to solve.

    Collinear = on the same straight line, meaning that the gradient joining all three lines points together, i.e. the gradient between any two points, is exactly the same (straight line equation is mx+c,where m is the gradient. Since they’re all on the same straight line, the value of the gradient must be the same, i.e. must always be m, between any two points). So because you’ve already worked out your gradient for AB, and you have an expression for the gradient of BC, but you know that for the points to be collinear these two values must be equal, so you can just equate them to each other. So gradient for AB=gradient for BC. Then you just solve it for y.

    Hope this helps, I’ve tried not to do all the sums because then it’s just copying down something you’ve read off tsr which isn’t much help long term. But yeah, hopefully this made more sense!! I definitely need sleep now but if you still don't get it then if someone else doesn't help you understand I'll try again tomorrow until you get it
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    (Original post by hhattiecc)
    Sorry for confusing you! Hopefully this will clear things up. So plot A and B on a graph. If you draw a line from A to B (i.e AB) then make a right angled triangle, you can work out the length of AB by using Pythagoras’ theorem with AB as the hypotenuse. So the difference in x values is one side length of your triangle, and the same with the y values (if you’re struggling to see this then look at your diagram). So difference in x values is 3-1=2. Difference in y values is 5-3=2. Using pythag: 2^2 + 2^2 =AB^2 = 8.
    You can do the same with AC but this time the difference in y values will be in terms of y. Difference in x=1- -1=2. Difference in y=3-y. Using pythag: 2^2 +(3-y)^2 = AC^2. AC=AB therefore AC^2 also equals 8. Hence you have an equation of 4 + (3-y)^2 = 8. This is just an equation you can solve for y, but you cansee that the answers of 1 and 5 are the solutions by subbing in those values.

    Take exactly the same approach with AC = BC. So (difference in x values)^2 +(difference in y values)^2 = (distance between two points)^2… I think this is a formula you need to know in A level. Should be in your course book I think.

    With the gradients… Change in y/change in x. So for finding the gradient of the line AB you should be able to do, and as for BC use the same equation: change in y =5-y, change in x=3- -1=4. So gradient for that is (5-y)/4. Perpendicular gradients have a product of -1 (again, just something you should know from your studies) so (gradient of AB) x (gradient ofBC) = -1. This will give you an algebraic equation in terms of y, which shouldbe pretty straight forward to solve.

    Collinear = on the same straight line, meaning that the gradient joining all three lines points together, i.e. the gradient between any two points, is exactly the same (straight line equation is mx+c,where m is the gradient. Since they’re all on the same straight line, the value of the gradient must be the same, i.e. must always be m, between any two points). So because you’ve already worked out your gradient for AB, and you have an expression for the gradient of BC, but you know that for the points to be collinear these two values must be equal, so you can just equate them to each other. So gradient for AB=gradient for BC. Then you just solve it for y.

    Hope this helps, I’ve tried not to do all the sums because then it’s just copying down something you’ve read off tsr which isn’t much help long term. But yeah, hopefully this made more sense!! I definitely need sleep now but if you still don't get it then if someone else doesn't help you understand I'll try again tomorrow until you get it
    This has definitely helped me, yes thanks but I'm still not getting the same answers as my text book, for iii) I managed to get 6, when the answer is 9. ii) is still a problem, although I followed the exact same steps for 'i'

    Here is how I done i so far)
    So AC^2 = 8

    BC = (x2-x1)^2 + (y2-y1)^2
    = (3--1)^2 + (5-y)^2
    = 4^2 + (5-y)^2

    From here I get very confused as to weather I need to expand the brackets so what? I hope I'm doing this correct to begin with.
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    (Original post by basketballgirl11)
    This has definitely helped me, yes thanks but I'm still not getting the same answers as my text book, for iii) I managed to get 6, when the answer is 9. ii) is still a problem, although I followed the exact same steps for 'i'

    Here is how I done i so far)
    So AC^2 = 8

    BC = (x2-x1)^2 + (y2-y1)^2
    = (3--1)^2 + (5-y)^2
    = 4^2 + (5-y)^2

    From here I get very confused as to weather I need to expand the brackets so what? I hope I'm doing this correct to begin with.
    Hi, what stage are you at now?

    For (i), are you happy that you get a quadratic equation for y which has 2 solutions, which is why the answer can be 1 or 5?

    For (ii), you're trying the right method but your AC^2 is wrong - it should be 4 + (y - 3)^2. Have another look at and see why this is the correct expression.

    Now when you expand the brackets and rearrange, something interesting will happen and you'll see that there can only be one value of y.

    For (iii) you can do it with gradients or the vector dot product if you've done vectors. Either way will give you the answer - post some working if still stuck
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    (Original post by basketballgirl11)
    This has definitely helped me, yes thanks but I'm still not getting the same answers as my text book, for iii) I managed to get 6, when the answer is 9. ii) is still a problem, although I followed the exact same steps for 'i'

    Here is how I done i so far)
    So AC^2 = 8

    BC = (x2-x1)^2 + (y2-y1)^2
    = (3--1)^2 + (5-y)^2
    = 4^2 + (5-y)^2

    From here I get very confused as to weather I need to expand the brackets so what? I hope I'm doing this correct to begin with.
    You've correctly worked out an expression for the length of BC squared, yes (remember you've got the (x2-x1)^2 + (y2-y1)^2 formula based on Pythagoras' theorem so it follows that this must be the expression for BC squared, not BC).

    The important thing to take into account here is that AC^2 is only equal to 8 in part i, not part ii. This is because A and B are fixed points, so the length of AB is always the same, whereas the length of AC and BC are completely variable depending on your value of y. When AB=AC as in i, then AC^2 =8 as AB^2=8. In part ii however, we are no longer considering the point C where AB=AC, so AC^2 is not equal to 8. In part ii you have to start fresh: so AC^2 is calculated in exactly the same way as you correctly calculated BC^2, i.e. by using (x2-x1)^2 + (y2-y1)^2. We know that AC=BC, and squaring both sides it follows that AC^2=BC^2. You've just worked out an expression for AC^2 and BC^2, so equate these. Yep, you're going to have to expand the brackets, and after you've done that and simplified it you should be able to get an equation you can solve for y - when you expand them your y^2 terms will cancel out, meaning you won't have a quadratic to solve and so you will only have one value of y as your solution.

    For iii I'm not quite sure how you got 6, if you post your working and then hopefully someone will be able to push you in the right direction if you still get to the wrong answer. For iii, just remember:
    Gradient= change in y / change in x = (y2-y1)/(x2-x1)
    If two straight lines are perpendicular, then the product of their gradients is equal to -1. So m1 x m2 = -1.
    If you work out the gradients of AB and BC using the formula above, and then put (AB gradient) x (BC gradient) = -1, you will get an expression in terms of y which you will be able to solve.

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    (Original post by hhattiecc)
    You've correctly worked out an expression for the length of BC squared, yes (remember you've got the (x2-x1)^2 + (y2-y1)^2 formula based on Pythagoras' theorem so it follows that this must be the expression for BC squared, not BC).

    The important thing to take into account here is that AC^2 is only equal to 8 in part i, not part ii. This is because A and B are fixed points, so the length of AB is always the same, whereas the length of AC and BC are completely variable depending on your value of y. When AB=AC as in i, then AC^2 =8 as AB^2=8. In part ii however, we are no longer considering the point C where AB=AC, so AC^2 is not equal to 8. In part ii you have to start fresh: so AC^2 is calculated in exactly the same way as you correctly calculated BC^2, i.e. by using (x2-x1)^2 + (y2-y1)^2. We know that AC=BC, and squaring both sides it follows that AC^2=BC^2. You've just worked out an expression for AC^2 and BC^2, so equate these. Yep, you're going to have to expand the brackets, and after you've done that and simplified it you should be able to get an equation you can solve for y - when you expand them your y^2 terms will cancel out, meaning you won't have a quadratic to solve and so you will only have one value of y as your solution.

    For iii I'm not quite sure how you got 6, if you post your working and then hopefully someone will be able to push you in the right direction if you still get to the wrong answer. For iii, just remember:
    Gradient= change in y / change in x = (y2-y1)/(x2-x1)
    If two straight lines are perpendicular, then the product of their gradients is equal to -1. So m1 x m2 = -1.
    If you work out the gradients of AB and BC using the formula above, and then put (AB gradient) x (BC gradient) = -1, you will get an expression in terms of y which you will be able to solve.

    You are getting close to putting a full solution which us not allowed. Next time please just give a hint or ask the OP to show what they've done.
 
 
 
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