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    Heres the question:

    The diagram below shows a circle with a centre ‘Z’.
    Line XY is a tangent to the circle at point ‘P’
    The coordinates of Z, X, and Y are (2, 3) (-6, -4) (8, 5) respectively.

    Find the equation of line ‘ZP’ and give it in the form ax+by+c=0

    Any help?
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    Ok so first of all, line ZP is perpendicular to line XY because a tangent touching the circumference of a circle means the radius is perpendicular to it. So, what we know about perpendicular lines is that they have the negative reciprocal. We can find this if we find the equation of the line XY.

    Now, to find the gradient ofXY, we simply divide the difference in Ys by the difference in Xs. This is simple as we have two co-ordinates: (-6,-4) and (8,5).

    So, the difference in Ys for these two co-ordinates is 9 (difference between -4 and 5).

    The difference between in the Xs is 14 (difference between -6 and 8).

    Therefore, the gradient of XY is 9/4. Leave it as a fraction (always do unless told so or it is necessary because it keeps things accurate). The incomplete equation is then y = 9/4x + C. We do not need to find the y intercept at the moment (I think).

    So from this equation of line XY is y = 9/4x + C, we can now find the gradient of the line ZP. Remember, ZP is perpendicular to XY so therefore ZP's gradient will be the negative reciprocal of XY. This basically means the fraction is flipped upside down and has a negative sign stuck on it!

    ZP = y = -4/9x + C

    To find the y intercept, substitute the Z co-ordinate into the equation, y = -4/9x + C. So for example, the Z co-ordinate is (2,3) which means x = 2 and y = 3. Rearrange the equation to find C. you should get that C is equal to 3.4 reoccurring.

    SO, the equation of the line ZP is y = -4/9 + 3.444...

    Honestly, I have no idea how to give it as a quadratic.
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    (Original post by BTAnonymous)
    Ok so first of all, line ZP is perpendicular to line XY because a tangent touching the circumference of a circle means the radius is perpendicular to it. So, what we know about perpendicular lines is that they have the negative reciprocal. We can find this if we find the equation of the line XY.

    Now, to find the gradient ofXY, we simply divide the difference in Ys by the difference in Xs. This is simple as we have two co-ordinates: (-6,-4) and (8,5).

    So, the difference in Ys for these two co-ordinates is 9 (difference between -4 and 5).

    The difference between in the Xs is 14 (difference between -6 and 8).

    Therefore, the gradient of XY is 9/4. Leave it as a fraction (always do unless told so or it is necessary because it keeps things accurate). The incomplete equation is then y = 9/4x + C. We do not need to find the y intercept at the moment (I think).

    So from this equation of line XY is y = 9/4x + C, we can now find the gradient of the line ZP. Remember, ZP is perpendicular to XY so therefore ZP's gradient will be the negative reciprocal of XY. This basically means the fraction is flipped upside down and has a negative sign stuck on it!

    ZP = y = -4/9x + C

    To find the y intercept, substitute the Z co-ordinate into the equation, y = -4/9x + C. So for example, the Z co-ordinate is (2,3) which means x = 2 and y = 3. Rearrange the equation to find C. you should get that C is equal to 3.4 reoccurring.

    SO, the equation of the line ZP is y = -4/9 + 3.444...

    Honestly, I have no idea how to give it as a quadratic.
    thanks, but the question is not asking it to give it as a quadratic, that is still a linear formaula
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    (Original post by theBranicAc)
    thanks, but the question is not asking it to give it as a quadratic, that is still a linear formaula
    https://www.youtube.com/watch?v=jHA8DtUmtDY

    What exam board are you doing btw?
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    (Original post by theBranicAc)
    thanks, but the question is not asking it to give it as a quadratic, that is still a linear formaula
    Take his gradient and Z coordinate as sub into:

    y-b = m(x-a)

    Rearrange to bring all terms to the left.
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    (Original post by BTAnonymous)
    https://www.youtube.com/watch?v=jHA8DtUmtDY

    What exam board are you doing btw?
    He asked for the answer in the form of the general equation for a straight line, not a quadratic.
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    (Original post by BTAnonymous)
    https://www.youtube.com/watch?v=jHA8DtUmtDY

    What exam board are you doing btw?
    edexcel
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    (Original post by theBranicAc)
    edexcel
    Is this GCSE?
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    (Original post by theBranicAc)
    edexcel
    Oh cool, I've never been taught how to convert this so I guess this is good revision for tomorrow.
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    (Original post by theBranicAc)
    Heres the question:

    The diagram below shows a circle with a centre ‘Z’.
    Line XY is a tangent to the circle at point ‘P’
    The coordinates of Z, X, and Y are (2, 3) (-6, -4) (8, 5) respectively.

    Find the equation of line ‘ZP’ and give it in the form ax+by+c=0

    Any help?
    First take the line xy and work out the gradient using y2-y2/ x2-x1. As the other user said, this gives a gradient of 9/14. Since XY is a tangent to the circle at point P, this means that Z joining up to P is perpendicular and consequently the gradient of ZP will be the negative reciprocal : -14/9. If you input the gradient and the co ordinates (2,3) into y=mx +c, you get 3= -14/9 x 2 +c and rearranged it gives you c as 6 and 55/9 so the equation is y= -14/9x + 55/9. Then, I assume you subtract y from the other side to get -14/9x - 1y -55/9 = 0.
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    (Original post by Speedbird129)
    Is this GCSE?
    yes it is
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    (Original post by theBranicAc)
    Heres the question:

    The diagram below shows a circle with a centre ‘Z’.
    Line XY is a tangent to the circle at point ‘P’
    The coordinates of Z, X, and Y are (2, 3) (-6, -4) (8, 5) respectively.

    Find the equation of line ‘ZP’ and give it in the form ax+by+c=0

    Any help?
    14x + 9y + 55 = 0
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    (Original post by BTAnonymous)
    Ok so first of all, line ZP is perpendicular to line XY because a tangent touching the circumference of a circle means the radius is perpendicular to it. So, what we know about perpendicular lines is that they have the negative reciprocal. We can find this if we find the equation of the line XY.

    Now, to find the gradient ofXY, we simply divide the difference in Ys by the difference in Xs. This is simple as we have two co-ordinates: (-6,-4) and (8,5).

    So, the difference in Ys for these two co-ordinates is 9 (difference between -4 and 5).

    The difference between in the Xs is 14 (difference between -6 and 8).

    Therefore, the gradient of XY is 9/4. Leave it as a fraction (always do unless told so or it is necessary because it keeps things accurate). The incomplete equation is then y = 9/4x + C. We do not need to find the y intercept at the moment (I think).

    So from this equation of line XY is y = 9/4x + C, we can now find the gradient of the line ZP. Remember, ZP is perpendicular to XY so therefore ZP's gradient will be the negative reciprocal of XY. This basically means the fraction is flipped upside down and has a negative sign stuck on it!

    ZP = y = -4/9x + C

    To find the y intercept, substitute the Z co-ordinate into the equation, y = -4/9x + C. So for example, the Z co-ordinate is (2,3) which means x = 2 and y = 3. Rearrange the equation to find C. you should get that C is equal to 3.4 reoccurring.

    SO, the equation of the line ZP is y = -4/9 + 3.444...

    Honestly, I have no idea how to give it as a quadratic.
    The gradient is actually -14/9
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    (Original post by theBranicAc)
    Heres the question:

    The diagram below shows a circle with a centre ‘Z’.
    Line XY is a tangent to the circle at point ‘P’
    The coordinates of Z, X, and Y are (2, 3) (-6, -4) (8, 5) respectively.

    Find the equation of line ‘ZP’ and give it in the form ax+by+c=0

    Any help?
    HI there, here are the solutions someone uploaded but the answer wasn't in quadratic for. Quadratics are ax^2+bx+c.
    Attached Files
  1. File Type: docx Detailed Worked Solutions to the 3 Questions.docx (291.3 KB, 62 views)
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    (Original post by theBranicAc)
    Heres the question:

    The diagram below shows a circle with a centre ‘Z’.
    Line XY is a tangent to the circle at point ‘P’
    The coordinates of Z, X, and Y are (2, 3) (-6, -4) (8, 5) respectively.

    Find the equation of line ‘ZP’ and give it in the form ax+by+c=0

    Any help?
    It's not asking you to give it as a quadratic.
    A quadratic would be y=ax^2+bx+c
    Here you just need to rearrange to put y on one side.

    ax+by+c = 0
    ax+c =by

    (a/b)x + (c/b) = y

    Then once you've got the equation of the line you can just convert it back to the original form.
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    Here's my working, not sure if it's right though.
    So feed back if anyone thinks I've went wrong.

    Attachment 546253

    EDIT: just realised I made a very stupid mistake lol, I will retry it the now.




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    (Original post by techfan42)
    The gradient is actually -14/9
    I don't know why I put -4. I even have it on my paper as -14. fml. The difference between -6 is 8 is definitely 4. God I feel stupid.
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    (Original post by offhegoes)
    He asked for the answer in the form of the general equation for a straight line, not a quadratic.
    That's what the video is about. I don't know myself how to convert it into the standard form equation so decided to find a video and share it with them
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    Ok here's my full working. I think it's right.

    Name:  ImageUploadedByStudent Room1465415054.738555.jpg
Views: 74
Size:  139.9 KB


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    What level is this ?


    Posted from TSR Mobile
 
 
 
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