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Help with coordinate Geometry

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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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gradient = up/across... you have done across/up :spank:
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
Reply 3
Avatar for TSR George
TSR George
OP
17
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 :/
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
Reply 5
Avatar for TSR George
TSR George
OP
17
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?
it crosses x axis when value of y=0 so you put value of y in and find out value of x from equation
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:
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
that formula is not compulsory... your method is fine.
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?
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?
Yeah i personally believe it is because you make less mistakes through steps
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?
Two coordinates
Original post by ElectronDonor
Two coordinates


thanks
Do you understand how to do rest?
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
Bump
hai! i need help with this question if anyone wants to help :smile:

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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