Hard Maths question

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 of XY, 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.

(edited 7 years ago)

Reply 2

theBranicAc

OP

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 of XY, 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

Reply 3

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?

Reply 4

1420787

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.

Reply 5

1420787

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.

Reply 6

theBranicAc

OP

Original post by BTAnonymous

https://www.youtube.com/watch?v=jHA8DtUmtDY

What exam board are you doing btw?

edexcel

Reply 7

Speedbird129

12

Original post by theBranicAc

edexcel

Is this GCSE?

Reply 8

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.

Reply 9

leopard923

11

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. :smile:

(edited 7 years ago)

Reply 10

theBranicAc

OP

Detailed Worked Solutions to the 3 Questions.docx298.3KB

Original post by Speedbird129

Is this GCSE?

yes it is

Reply 11

techfan42

4

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

Reply 12

techfan42

4

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 of XY, 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

Reply 13

username2437011

13

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.

Reply 14

username2207531

15

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.

Reply 15

username2548927

18

Here's my working, not sure if it's right though.
So feed back if anyone thinks I've went wrong.

Attachment not found

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

Posted from TSR Mobile

(edited 7 years ago)

Reply 16

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.

Reply 17