core 1 coordinate geometry help!!! Watch

shana_t
Badges: 1
Rep:
?
#21
Report Thread starter 5 years ago
#21
(Original post by Robbie242)
You have two points A(1, 3\sqrt3), B (2+\sqrt3,3+4\sqrt3) You've being told that the gradient of the line l which passes through these two points is m=\sqrt{3}

Now the equation of a line is given by the formula y-y_{1}=m(x-x_{1}) This means that you can substitute any of the coordinates from either point to get the equation of l, so lets label A as the point A(x_{1},y_{1}) well from the question x_{1}=1 and y_{1}=3\sqrt3 for the coordinates of A correct?

Substitute these into the formula along with m, expand and rearrange into the form y=mx+c
thnx
0
quote
reply
shana_t
Badges: 1
Rep:
?
#22
Report Thread starter 5 years ago
#22
(Original post by BabyMaths)
Here's a simpler example. A line with gradient 2 passes through the point (4,9). Find the equation of the line in the form y=mx+c<br />
<br />
You already know the equation is y=2x+c and you know that y=9 when x=4. <br />
<br />
You can use those values to find the value of c.<br />
<br />
9=2 \times 4 + c<br />
<br />
c=1<br />
<br />
y=2x+1
<br />
<br />

thnx
0
quote
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

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

Personalise

University open days

  • Sheffield Hallam University
    City Campus Undergraduate
    Thu, 13 Dec '18
  • University of Buckingham
    Open Evening Undergraduate
    Thu, 13 Dec '18
  • University of Lincoln
    Mini Open Day at the Brayford Campus Undergraduate
    Wed, 19 Dec '18

Were you ever put in isolation at school?

Yes (18)
31.03%
No (40)
68.97%

Watched Threads

View All