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    Please...Help me...
    I have to find the area of PQR. I found the co-ordinates of P which are (0, 1).
    I know I must use Pythagora's theorem but I am not getting the right answer. I am getting 6 since I did half times base (2) times height (6).
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    (Original post by CounTolstoy)
    Please...Help me...
    I have to find the area of PQR. I found the co-ordinates of P which are (0, 1).
    I know I must use Pythagora's theorem but I am not getting the right answer. I am getting 6 since I did half times base (2) times height (6).
    Find the coordinates of the x-intercept of the line l2.

    Then you'll be able to work out the base and height of the triangle if you consider the coodinates Q, R and the x-intercept of l2.
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    (Original post by CounTolstoy)
    I have to find the area of PQR. I found the co-ordinates of P which are (0, 1).
    I agree with this.
    (Original post by CounTolstoy)
    I know I must use Pythagora's theorem but I am not getting the right answer. I am getting 6 since I did half times base (2) times height (6).
    How do you get those values for the base and the height?
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    (Original post by notnek)
    Find the coordinates of the x-intercept of the line l2.

    Then you'll be able to work out the base and height of the triangle if you consider the coodinates Q, R and the x-intercept of l2.
    I found the equation of l2 which is y = 2x + 1.
    My point is, I have all the information I need and I am aware that to find the area I have to do 1/2 times PQ times QR. However, in the mark cheme, when they are finding PQ, they use the method as though they're calculating the hypotenuse of the smaller triangle rather than the opposite or adjacent side. Because hypotenuse squared = A^2 + B^2. But as you can see, the length of PQ isn't the hypotenuse of the smaller triangle so shouldn't it be Hypotenuse squared minus adjacent squared to give PQ squared?
    Sorry, I'm not sure if I have explained this very well.
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    (Original post by Pangol)
    I agree with this.

    How do you get those values for the base and the height?
    The base can be obtained by using Pythagoras theorem. The base is PQ so I would do hypotenuse squared (of smaller triangle) minus the small length between point Q and the y axis squared. Then I would square root this for PQ. Thus: 2 squared - 1 squared = 3. So PQ = square root of three. But in the mark scheme, it's adding 2 squared to 1 squared. Is this not incorrect since one would only do that if the hypotenuse is being calculated?
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    The base and height you need are the lengths of the sides PQ and PR.

    You have already shown in part (a) that the length of PR is not 6. Whatever you did there, use the same idea to find the length of PQ (which is not 2).
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    (Original post by Pangol)
    The base and height you need are the lengths of the sides PQ and PR.

    You have already shown in part (a) that the length of PR is not 6. Whatever you did there, use the same idea to find the length of PQ (which is not 2).
    PQ = Square root of 3. Is this correct? Mark Scheme says square root of 5.
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    (Original post by CounTolstoy)
    PQ = Square root of 3. Is this correct? Mark Scheme says square root of 5.
    The mark scheme is correct. How are you working out this distance? And what did you get for QR?
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    (Original post by Pangol)
    The mark scheme is correct. How are you working out this distance? And what did you get for QR?
    Damn, I was so sure it was wrong.
    QR = 3 root five according to me...
    For PQ, I am doing 2 squared minus 1 squared...since 2 squared is the hypotenuse (we are moving from (0, 1) to (1,3) and the other smaller side is 1 as the x co-ordinate increase by one: 0 ---> 1
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    I'm afraid that you are working out your distances in the wrong way. When you say "2 squared is the hypotenuse", which right angled triangle are you picturing?

    Your QR is right - try using the method you used to get that.
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    (Original post by Pangol)
    I'm afraid that you are working out your distances in the wrong way. When you say "2 squared is the hypotenuse", which right angled triangle are you picturing?

    Your QR is right - try using the method you used to get that.
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    The hypotenuse I am referring to is the side labelled Z in my diagram. Now I am doubting whether the triangle PQR I am attempting to find is indeed the one I have tried to depict (the one with the drawn right angle). I hope my photo is helping to clarify where I've gone wrong because I've no idea! Thank you in advance...
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    Your diagram reveals your problem - the length of side Z is bigger than 2, because the top of side Z is higher up than Q.

    Draw a horizontal line across from P. Draw a vertical line down from Q. The triangle formed by P, Q, and the point of intersection of these two lines is the right angled triangle you want to use.
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    (Original post by Pangol)
    Your diagram reveals your problem - the length of side Z is bigger than 2, because the top of side Z is higher up than Q.

    Draw a horizontal line across from P. Draw a vertical line down from Q. The triangle formed by P, Q, and the point of intersection of these two lines is the right angled triangle you want to use.
    Ahhh yes! I see what triangle I should have used. Thank you so much for your help. I just have a final question: is there any way I could have known which triangle to use?
    Thank you so much, I fully appreciate your time and help.
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    In any question where you have two points, (x1,y1) and (x2,y2), and you want to find the distance between them, the triangle will be one like I describe above. The hypotenuse is always the distance between the points. There is an equation that goes with this;

    (square root of) (x1-x2)^2 + (y1-y2)^2,

    but that equation is saying the same thing as "find the length of the hypotenuse of the right angled triangle above".

    If there are already lines and/or triangles drawn in the question, ignore them and focus on this.
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    (Original post by Pangol)
    In any question where you have two points, (x1,y1) and (x2,y2), and you want to find the distance between them, the triangle will be one like I describe above. The hypotenuse is always the distance between the points. There is an equation that goes with this;

    (square root of) (x1-x2)^2 + (y1-y2)^2,

    but that equation is saying the same thing as "find the length of the hypotenuse of the right angled triangle above".

    If there are already lines and/or triangles drawn in the question, ignore them and focus on this.
    Thank you so much, Pangol. I truly appreciate all the help you've given me!!!
 
 
 
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