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# C1 help pls again heh watch

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1. How did they do part iv

Attachment 608146
2. I can't view the image
3. (Original post by CorpusLuteum)
How did they do part iv

Attachment 608146
4. which year is it ?
5. (Original post by h3rmit)
(Original post by whitedwarf)
I can't view the image
I attached it again
6. (Original post by mgy2000)
which year is it ?
2014 june OCR MEI
7. First find the coordinates for d. Then use y-y1=m(x-x1) by subbing in A and D's coordinates and then you can find m. Then sub in m and d coordinates into y=mx+c and find c. Final form will be y= (your m)x+(Your c)
8. work out the gradient of bc
flip the fraction and put a negative as ad is the negative reciprocal of bc , this will give u gradient of ad
y-y1=m(x-x1) where m is gradient and sub the coordinates of c and there u hv it
9. (Original post by CorpusLuteum)
...
Find the gradient of the line AD and call this m. Then the gradient of the tangent line at D is the negative reciprocal of m.
10. (Original post by CorpusLuteum)
I attached it again
Find the gradient of AC or AD or DC, and use that in y=mx+c
11. (Original post by h3rmit)
Find the gradient of AC or AD or DC, and use that in y=mx+c
but why?
12. (Original post by CorpusLuteum)
but why?
So you can pass your maths exams, get into a good uni, get a good job and be prepared for life? I'm not really sure what justifcation you're looking for
13. I do OCR without the MEI so idk if you would do it the same way as me but I hope this helps

Find the gradient of the line AD as this will be perpendicular to the tangent that you need to find.
(4-0)/(7-1) = 2/3
The gradient of the tangent will be the negative reciprocal of this. (Because they are perpendicular to each other)
-3/2
Put this gradient into a straight line equation
y=mx+c
y=-3/2x+c
Substitute your x and y for D in to work out c
4=-3/2(7) + c
c=14.5

So y=(-3/2)x+14.5
14. Work out the gradient of AD. The gradient of the tangent at D is the negative reciprocal of this since you know that a tangent is perpendicular to a diameter (or radius). From there you should be able to work out the equation of the line
15. (Original post by h3rmit)
So you can pass your maths exams, get into a good uni, get a good job and be prepared for life? I'm not really sure what justifcation you're looking for
lol, i meant why find the gradient of AC isn't a tangent perpendicular to it?
16. (Original post by CorpusLuteum)
lol, i meant why find the gradient of AC isn't a tangent perpendicular to it?
Yes
17. (Original post by klm7)
I do OCR without the MEI so idk if you would do it the same way as me but I hope this helps

Find the gradient of the line AD as this will be perpendicular to the tangent that you need to find.
(4-0)/(7-1) = 2/3
The gradient of the tangent will be the negative reciprocal of this. (Because they are perpendicular to each other)
-3/2
Put this gradient into a straight line equation
y=mx+c
y=-3/2x+c
Substitute your x and y for D in to work out c
4=-3/2(7) + c
c=14.5

So y=(-3/2)x+14.5
tenks i get it

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