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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 1
 08062016 19:24

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 2
 08062016 19:53
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 coordinates: (6,4) and (8,5).
So, the difference in Ys for these two coordinates 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 coordinate into the equation, y = 4/9x + C. So for example, the Z coordinate 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.Last edited by BTAnonymous; 08062016 at 19:56. 
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 3
 08062016 20:10
(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 coordinates: (6,4) and (8,5).
So, the difference in Ys for these two coordinates 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 coordinate into the equation, y = 4/9x + C. So for example, the Z coordinate 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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 4
 08062016 20:14
(Original post by theBranicAc)
thanks, but the question is not asking it to give it as a quadratic, that is still a linear formaula
What exam board are you doing btw? 
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 5
 08062016 20:14
(Original post by theBranicAc)
thanks, but the question is not asking it to give it as a quadratic, that is still a linear formaula
yb = m(xa)
Rearrange to bring all terms to the left. 
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 6
 08062016 20:15
(Original post by BTAnonymous)
https://www.youtube.com/watch?v=jHA8DtUmtDY
What exam board are you doing btw? 
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 7
 08062016 20:16
(Original post by BTAnonymous)
https://www.youtube.com/watch?v=jHA8DtUmtDY
What exam board are you doing btw? 
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 8
 08062016 20:18
(Original post by theBranicAc)
edexcel 
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 9
 08062016 20:18
(Original post by theBranicAc)
edexcel 
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 10
 08062016 20:20
(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?Last edited by leopard923; 08062016 at 20:22. 
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 11
 08062016 20:22
(Original post by Speedbird129)
Is this GCSE? 
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 12
 08062016 20:22
(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? 
 Follow
 13
 08062016 20:24
(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 coordinates: (6,4) and (8,5).
So, the difference in Ys for these two coordinates 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 coordinate into the equation, y = 4/9x + C. So for example, the Z coordinate 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. 
 Follow
 14
 08062016 20:24
(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? 
 Follow
 15
 08062016 20:26
(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?
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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 16
 08062016 20:34
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.
Posted from TSR MobileLast edited by RossB1702; 08062016 at 20:37. 
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 17
 08062016 20:35
(Original post by techfan42)
The gradient is actually 14/9 
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 18
 08062016 20:42
(Original post by offhegoes)
He asked for the answer in the form of the general equation for a straight line, not a quadratic. 
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 19
 08062016 20:44
Ok here's my full working. I think it's right.
Posted from TSR Mobile 
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 20
 08062016 20:45
What level is this ?
Posted from TSR Mobile
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Updated: June 8, 2016
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