# C1 help pls again hehWatch

Announcements
#1
How did they do part iv

Attachment 608146
0
2 years ago
#2
I can't view the image
0
#3
0
2 years ago
#4
(Original post by CorpusLuteum)
How did they do part iv

Attachment 608146
0
2 years ago
#5
which year is it ?
0
#6
(Original post by h3rmit)
(Original post by whitedwarf)
I can't view the image
I attached it again
0
#7
(Original post by mgy2000)
which year is it ?
2014 june OCR MEI
0
2 years ago
#8
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)
0
2 years ago
#9
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
0
2 years ago
#10
(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.
1
2 years ago
#11
(Original post by CorpusLuteum)
I attached it again
Find the gradient of AC or AD or DC, and use that in y=mx+c
0
#12
(Original post by h3rmit)
Find the gradient of AC or AD or DC, and use that in y=mx+c
but why?
0
2 years ago
#13
(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
1
2 years ago
#14
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
1
2 years ago
#15
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
2
#16
(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?
0
2 years ago
#17
(Original post by CorpusLuteum)
lol, i meant why find the gradient of AC isn't a tangent perpendicular to it?
Yes
0
#18
(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
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Bournemouth University
Midwifery Open Day at Portsmouth Campus Undergraduate
Wed, 16 Oct '19
• Teesside University
Wed, 16 Oct '19
• University of the Arts London
London College of Fashion – Cordwainers Footwear and Bags & Accessories Undergraduate
Wed, 16 Oct '19

### Poll

Join the discussion

#### How has the start of this academic year been for you?

Loving it - gonna be a great year (131)
18.3%
It's just nice to be back! (194)
27.09%
Not great so far... (257)
35.89%
I want to drop out! (134)
18.72%