Higher maths Straight line Help [Archive]

PDA

View Full Version : Higher maths Straight line Help

BowTies1

28-08-2008, 09:58 PM

The Professor

28-08-2008, 10:06 PM

Do you want the answers or how to work them out?

BowTies1

28-08-2008, 10:08 PM

a god walkthrew on how to work them out would be fine that is just like a a few out oflikie 30 ive to do so answers woudnt bother me either cause im trying more so this is jsut like examples so answers or a good walkthrew is fine

edit : i dont want JUST the answers i need to fiqure out how to do it

kk.

28-08-2008, 10:19 PM

BowTies1

28-08-2008, 10:21 PM

my sheet says it so it must be soemthing differnt you do

kk.

28-08-2008, 10:25 PM

trust me, it doesnt work. read the edit. ill see if i can be bothered to do the other ones

BowTies1

28-08-2008, 10:35 PM

ok thanks would be a great help

kk.

28-08-2008, 10:44 PM

ok
5 b) center of line PQ is (-1,6) so..

m1m2 = -1 ............ m1 = 1/3

1/3 x m2 = -1
m2 = -1/.333
m2 = -3

y - y1 = m(x - x1)
y - 6 = -3 (x - -1)
y = -3 (x + 1) + 6
y = -3x -3 + 6

y = -3x + 3

substitute in (-1,6)

the last one doesnt make sense to me

Jesus-Egg

28-08-2008, 11:50 PM

4) Doesnt work. 2 - 12 + 7 doesnt = 0

Read the question. (2,3) lies on the perpendicular, not on the line x - 4y + 7 = 0.

kk.

29-08-2008, 03:07 PM

Read the question. (2,3) lies on the perpendicular, not on the line x - 4y + 7 = 0.

Don't correct me until u read the question. It doesn't say lies, it's the point on that line. U need the point 2,3 to work for both otherwise u can't work it out. I just finihed my AS course and I know what I'm on about

Jesus-Egg

29-08-2008, 10:39 PM

Don't correct me until u read the question. It doesn't say lies, it's the point on that line. U need the point 2,3 to work for both otherwise u can't work it out. I just finihed my AS course and I know what I'm on about

I have read the question, and I got an A in A Level maths, and an A in AS further maths, so I know what I'm on about. What you did didn't work because x - 4y + 7 doesn't pass through the point (2,3). It is the perpendicular to this line that does.

You are finding one of the lines which is perpendicular to the line x - 4y + 7 = 0. The gradient of this line is 1/4, therefore the gradient of the perpendicular line must be -4. This perpendicular line also goes through the point (2,3). y = -4x + c, sub in (2,3), and find c.

Ostinato

29-08-2008, 10:53 PM

Don't correct me until u read the question. It doesn't say lies, it's the point on that line. U need the point 2,3 to work for both otherwise u can't work it out. I just finihed my AS course and I know what I'm on about

Well I have read the question and you seriously need corrected, because for you to say that "it doesn't work" considering you just finished your course is pretty atrocious tbh.

I got an A in Higher Maths and that is a very simple straight line question tbh.

To find the equation of a straight line you need a point and the gradient. In this case it has given you the point already (2,3). Therefore you need to work out the gradient next.

As you are told it is perpendicular to the line x - 4y + 7 = 0 you therefore can conclude that it has the same gradient except made perpendicular.

THEREFORE, you first of all work out the gradient of x - 4y + 7 = 0. To do this you need to have the equation in the form y= .......

SO-
x - 4y + 7 = 0
x + 7 + 4y
4y = x + 7

You then take the 4 over to get y on it's own...

y = 1/4x + 7

And using the rule that y=mx + c (with m being the gradient) you know that the gradient of this line is 1/4... However you need to make this perpendicular to get the answer we wish so the gradient you use is therefore -4

Because of this, you now have the point (2,3) and your gradient -4 so you can answer the question and work out the equation of the line using the standard formula y - b = m (x - a) =>

y - 3 = -4 (x - 2)
y - 3 = -4x +8
y = -4x + 11

SO PLEASE do not come barging in this thread thinking your some sort of know it all when quite clearly you don't just because you've sat your course. I've passed my Higher maths exam and I'm studying accounts at uni, so shush up.

kk.

30-08-2008, 03:56 PM

I have read the question, and I got an A in A Level maths, and an A in AS further maths, so I know what I'm on about. What you did didn't work because x - 4y + 7 doesn't pass through the point (2,3). It is the perpendicular to this line that does.

You are finding one of the lines which is perpendicular to the line x - 4y + 7 = 0. The gradient of this line is 1/4, therefore the gradient of the perpendicular line must be -4. This perpendicular line also goes through the point (2,3). y = -4x + c, sub in (2,3), and find c.
yeh, ive re read it for a final time and i now agree that it isnt on the line x-4y etc... It is a slightly mis leading question to what the papers normally set (for me anyway)

Well I have read the question and you seriously need corrected, because for you to say that "it doesn't work" considering you just finished your course is pretty atrocious tbh.

I got an A in Higher Maths and that is a very simple straight line question tbh.

To find the equation of a straight line you need a point and the gradient. In this case it has given you the point already (2,3). Therefore you need to work out the gradient next.

As you are told it is perpendicular to the line x - 4y + 7 = 0 you therefore can conclude that it has the same gradient except made perpendicular.

THEREFORE, you first of all work out the gradient of x - 4y + 7 = 0. To do this you need to have the equation in the form y= .......

SO-
x - 4y + 7 = 0
x + 7 + 4y
4y = x + 7

You then take the 4 over to get y on it's own...

y = 1/4x + 7

And using the rule that y=mx + c (with m being the gradient) you know that the gradient of this line is 1/4... However you need to make this perpendicular to get the answer we wish so the gradient you use is therefore -4

Because of this, you now have the point (2,3) and your gradient -4 so you can answer the question and work out the equation of the line using the standard formula y - b = m (x - a) =>

y - 3 = -4 (x - 2)
y - 3 = -4x +8
y = -4x + 11

SO PLEASE do not come barging in this thread thinking your some sort of know it all when quite clearly you don't just because you've sat your course. I've passed my Higher maths exam and I'm studying accounts at uni, so shush up.
can i just say, even though you got an A and i would have though, even though it makes no difference to the answer, that you would have divided 7 by 4.

but that last bit wasnt needed... I helpe him on 2 or 3 of the questions before you even posted in the thread and accounts... big whoop, im planning on doing accounting and finance and im pretty sure that decision maths is much more important (or so what i was told by one of my teachers)

its funny because i did start of how u did, it just didnt click to use the y -y1 = m( x - x1)

Emzy

30-08-2008, 06:55 PM

o.o
****
If thats higher maths i am completely screwed.

Soka

30-08-2008, 07:55 PM

find the equation of the line through the point (3,5) which is parallel to the line with the equation 3x+2y -5 = 0

4) find the equation of the line through the point (2,3) prependicular to the line x-4y +7 = 0

5) P & are the points (-4,5) and (2,7)
find the equation of

(a) The line PQ
(B) The perpendicular Bisector of PQ

6. find the equation of the median Ad of traingle ABC where the coordinates of A,B and c are (-2,3) (-3,-4) and (5,2) respectively

lol at the arguing over basic co-ordinate geomety

i dont know if all being answered yet so ill answer them all and try to explain how to come about it as im bored :P

3) 3x + 2y - 5 = 0
we need gradient of this so simple rearranging
2y = 5 - 3 x
y = 5/2 - 3/2x
and putting in form y = mx+c, m is gradient therefore gradient = -3/2
parallel line = same gradient, so using
y = mx + c and subbing in (3,5)
5 = (-3/2 x 3) + c
5 = -9/2 + 19/2
thus equation answer is y = -3/2x + 19/2 and tidying up to give integers
2y = 19 - 3x

4) x - 4y + 7 = 0
again, need gradient
x + 7 = 4y
x/4 + 7/4 = y
using y = mx+c gradient , m, is therefore 1/4
perpendicular line = negative reciprocal gradient, so gradient of equation were finding will be -4.
tip: if the two gradients multiply to minus 1 you'll know their perpendicular
using y = mx+c and subbing in (2,3)
3 = (2x-4) + c
3 = -8 + 11
thus equation is y = -4x + 11

5)
a) to find equation we need gradient first, of the two points
using simple dy/dx = (7-5)/ (2--4) = 2/6
then we use y = mx + c and sub. in any one of the points p or q
using q,
7 = (2 x 2/6) + c
7 = 4/6 + 38/6
y = 2/6x + 38/6, and tidying up
thus equation is 6y = 2x + 38

b) if im correct from memory bisector is through midpoint, so we need to find midpoint of p and q, which you simply add (x1 + x2)/2 and (y1 + y2)/2
to get midpoint, so midpoint is (-1, 6)
if its perpendicular gradient will be negative reciprocal of original equation gradient (2/6), so gradient of this is -6/2 = -3
again using y = mx + c and subbing in (-1,6) - point it passes through
6 = (-1 x -3) + c
6 = 3 + 3
thus equation is y = -3x + 3

with 6, i don't understand what it's asking for - you sure you've written it out right? let me know what it wants and i'll try help you do it :D

hope other questions help / are right - anyone correct me if i'm wrong

jesus-egg what modules you do for further maths? (i'm thinking of doing AS in further maths, alongside my A2 in normal maths - just achieved A in as)

Ostinato

30-08-2008, 11:21 PM

yeh, ive re read it for a final time and i now agree that it isnt on the line x-4y etc... It is a slightly mis leading question to what the papers normally set (for me anyway)

can i just say, even though you got an A and i would have though, even though it makes no difference to the answer, that you would have divided 7 by 4.

but that last bit wasnt needed... I helpe him on 2 or 3 of the questions before you even posted in the thread and accounts... big whoop, im planning on doing accounting and finance and im pretty sure that decision maths is much more important (or so what i was told by one of my teachers)

its funny because i did start of how u did, it just didnt click to use the y -y1 = m( x - x1)

No, it is a common mistake that people think you divide the 7 by the 4, but you don't lol...

Sorry if I seemed harsh but I was just a bit peed off with your post, it was very high and mighty about how you have just finished your course so you must be right and all this crap, when quite clearly you were wrong. lol

Obviously it is kind that you helped him in the other questions though and I am sure he appreciates your help.

kk.

31-08-2008, 05:33 PM

No, it is a common mistake that people think you divide the 7 by the 4, but you don't lol...

Sorry if I seemed harsh but I was just a bit peed off with your post, it was very high and mighty about how you have just finished your course so you must be right and all this crap, when quite clearly you were wrong. lol

Obviously it is kind that you helped him in the other questions though and I am sure he appreciates your help.
im sure you do divide the 7? its only when theyre in brackets or multiplied or divided that you divide it like (5x5)/7

i didnt mean to come across like that.

Jesus-Egg

31-08-2008, 09:34 PM

can i just say, even though you got an A and i would have though, even though it makes no difference to the answer, that you would have divided 7 by 4.

No, it is a common mistake that people think you divide the 7 by the 4, but you don't lol...

im sure you do divide the 7? its only when theyre in brackets or multiplied or divided that you divide it like (5x5)/7

Yes, you do divide the 7. 4y = x + 7 and to get y on its own, you need to remember to divide all of the rhs by 4, so you get y = (x + 7) / 4, which can be split up into y = (1/4)x + 7/4

jesus-egg what modules you do for further maths? (i'm thinking of doing AS in further maths, alongside my A2 in normal maths - just achieved A in as)

You have to do FP1, but you can choose the other 2 modules. I did D1 and M2. FP1 is fairly straightforward as long as you're good at the calculus

Lestah

31-08-2008, 10:45 PM

This AS-Level? I'm just leaving GCSE and thought that the starting bit was pretty easy

BowTies1

02-09-2008, 05:56 PM

This AS-Level? I'm just leaving GCSE and thought that the starting bit was pretty easy this is Scottish higher level i do not know what the equivalent is in england

thanks to everyone that helped me much appreciated

kk.

02-09-2008, 06:00 PM

yes it's an AS level question but it gets a bit harder

Want to hide these adverts? Register an account for free!