ihatePE
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I want to do maths A levels and i'm doing the step up course now, everything is going well until my teacher gave me some exam style question on this topic and feel like not taking maths after all. it's stressing me that i dont know how to do it.

Im making this thread to post questions that im not sure of, please help check/show me steps etc thanks!


1st question is:
Find equation of straight line that passes through the points (3, -1) and (-2, 2)
give your answer in form of ax+by+c=0
hence find the coordinates of the mid-point of intersection of the line and the x-axis


what i got so far is you find the gradient
-2-3 = -5
2 - -1 = 3
so -5/3 which i've put into y=-5/3x + c
i substitute a point in 2 = -5/3 * -2 + c
which becomes 2 = -10/3 + c
and now im stuck
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the bear
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gradient = up/across... you have done across/up :spank:
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ElectronDonor
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(Original post by ihatePE)
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I want to do maths A levels and i'm doing the step up course now, everything is going well until my teacher gave me some exam style question on this topic and feel like not taking maths after all. it's stressing me that i dont know how to do it.

Im making this thread to post questions that im not sure of, please help check/show me steps etc thanks!


1st question is:
Find equation of straight line that passes through the points (3, -1) and (-2, 2)
give your answer in form of ax+by+c=0
hence find the coordinates of the mid-point of intersection of the line and the x-axis


what i got so far is you find the gradient
-2-3 = -5
2 - -1 = 3
so -5/3 which i've put into y=-5/3x + c
i substitute a point in 2 = -5/3 * -2 + c
which becomes 2 = -10/3 + c
and now im stuck
gradient is change in y over change in x so (-1--2)/(3--2)=3/5=m
so y-y1=m(x-x1) so y+1=3/5(x-3) which equals 5y-3x+14=0
Midpoint is (x1+x2)/2,(y1+y2)/2 so (3+-2)/2,(-1+2)/2 so 0.5,0.5
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ihatePE
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(Original post by the bear)
gradient = up/across... you have done across/up :spank:
oh...

y=-3/5 x + c
2 = -3/5 *-2 + c
2 = 6/10 + c
now what :/
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ElectronDonor
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(Original post by ihatePE)
oh...

y=-3/5 x + c
2 = -3/5 *-2 + c
2 = 6/10 + c
now what :/
if your doing a level maths then your doing wrong equation formula
use this one
y-y1=m(x-x1) where (x1,y1) is a coordinate and m is gradient
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ihatePE
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(Original post by ElectronDonor)
if your doing a level maths then your doing wrong equation formula
use this one
y-y1=m(x-x1) where (x1,y1) is a coordinate and m is gradient
i havent got taught that formula T_T what does this equation help you find in general?
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(Original post by ihatePE)
oh...

y=-3/5 x + c
2 = -3/5 *-2 + c
2 = 6/10 + c
now what :/
it crosses x axis when value of y=0 so you put value of y in and find out value of x from equation
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the bear
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(Original post by ihatePE)
oh...

y=-3/5 x + c
2 = -3/5 *-2 + c
2 = 6/10 + c
now what :/
just subtract 0.6 from 2 to find c...

then write

y = -3/5x + your value of c

remember they don't want fractions, so you will need to multiply through by a number.

this method is perfectly fine for A level, despite what other people ^^^ may say :teehee:
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ElectronDonor
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(Original post by ihatePE)
i havent got taught that formula T_T what does this equation help you find in general?
it help you find the equation of a line, it is the one you a taught in a level maths, do you know what examination board your college/sixth form does for maths
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the bear
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(Original post by ihatePE)
i havent got taught that formula T_T what does this equation help you find in general?
that formula is not compulsory... your method is fine.
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ihatePE
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(Original post by ElectronDonor)
it help you find the equation of a line, it is the one you a taught in a level maths, do you know what examination board your college/sixth form does for maths
We do WJEC at A-level maths , so that equation is like y=mx + c but would be more helpful in cases like this right?
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ihatePE
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(Original post by the bear)
just subtract 0.6 from 2 to find c...

then write

y = -3/5x + your value of c

remember they don't want fractions, so you will need to multiply through by a number.

this method is perfectly fine for A level, despite what other people ^^^ may say :teehee:
ok from then i got y = -3/5x + 1.4
do i rearrange now to get it into the format they want?
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ElectronDonor
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(Original post by ihatePE)
We do WJEC at A-level maths , so that equation is like y=mx + c but would be more helpful in cases like this right?
Yeah i personally believe it is because you make less mistakes through steps
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ihatePE
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(Original post by ElectronDonor)
Yeah i personally believe it is because you make less mistakes through steps
when do you recommend using this formula? when they give me one points or two points or what type of questions?
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(Original post by ihatePE)
when do you recommend using this formula? when they give me one points or two points or what type of questions?
Two coordinates
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ihatePE
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(Original post by ElectronDonor)
Two coordinates
thanks
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(Original post by ihatePE)
thanks
Do you understand how to do rest?
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ihatePE
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(Original post by ElectronDonor)
Do you understand how to do rest?
im doing it later, im going to do some practise with this equation first, need to get it in my head before i move on thanks
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ihatePE
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hai! i need help with this question if anyone wants to help

Point A is (1,7) and B is (3,1)
midpoint of AB is P.
find the equation of the straight line which passes through P and which is perpendicular to the line 5y +x=7


i've found P has coordinate of (2,4) using the midpoint rule,
i also did equation of AB which is y = -3x + 10
i've found one of the coordinates for equation 5y + x = 7 and it's (7,1.4)

but then im stuck on what to do next with all these discoveries...any clue?
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ElectronDonor
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(Original post by ihatePE)
Bump
hai! i need help with this question if anyone wants to help

Point A is (1,7) and B is (3,1)
midpoint of AB is P.
find the equation of the straight line which passes through P and which is perpendicular to the line 5y +x=7


i've found P has coordinate of (2,4) using the midpoint rule,
i also did equation of AB which is y = -3x + 10
i've found one of the coordinates for equation 5y + x = 7 and it's (7,1.4)

but then im stuck on what to do next with all these discoveries...any clue?
Findv the gradient of the perpendicular line, a perpendicular line has negative reciprocal of line
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