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# core 1 coordinate geometry help!!! watch

1. cn anyone plz help how to work out the following qs:
1. A(-1,-2) B(7,2) and C(k,4), where k is constant, are the vertices of triangle ABC.
Angle ABC is a right angle. the gradient of AB is 1/2x
a) calculate the value of k

I hv no clue
2. (Original post by shana_t)
cn anyone plz help how to work out the following qs:
1. A(-1,-2) B(7,2) and C(k,4), where k is constant, are the vertices of triangle ABC.
Angle ABC is a right angle. the gradient of AB is 1/2x
a) calculate the value of k

I hv no clue
The gradient of AB = 1/2.

AB is perpendicular to BC.

What is the gradient of BC?

3. My way may not be the best but it seems to work for this question:

The gradient of AB= 1/2 and you said that it is a right angle triangle so you do the perpendicular so m2= -2

i used the coordinates B(7,2) and C(K,4)

rearrange:
2=-2(k-7) 2=-2K+14 2K=12 K=6

Hope that is explained well.
4. thnx for ur help, I got it
5. (Original post by shana_t)
cn anyone plz help how to work out the following qs:
1. A(-1,-2) B(7,2) and C(k,4), where k is constant, are the vertices of triangle ABC.
Angle ABC is a right angle. the gradient of AB is 1/2x
a) calculate the value of k

I hv no clue
As we have a right angle triangle, AB and BC perpendicular to one another. You have the gradient of AB as 1/2. You can work out the gradient of the perpendicular as 1/2 X m=-1 so m=-2 Now you know that the perpendicular to AB which is BC is a line with gradient -2. The equation of AC needs to be worked out. This is because you need the x value (k) at which y=4. You know that AC has gradient -2 and passes through point A, whose coordinates you already have as (-1, -2)
6. sorry again, I got stuck on this qs too:

the straight line l1 passes through the points A and B with coordinates (0,-2) and (6,7)
the equation of l1 is y=3/2x-2
another straight line l2 with equation x+y=8 cuts the y-axis at point . The lines l1 and l2 intersect at point D (4,4)
calculate the area of triangle ACD
7. (Original post by krisshP)
As we have a right angle triangle, AB and AC are perpendicular to one another. You have the gradient of AB as 1/2. You can work out the gradient of the perpendicular as 1/2 X m=-1 so m=-2 Now you know that the perpendicular to AB which is AC is a line with gradient -2. The equation of AC needs to be worked out. This is because you need the x value (k) at which y=4. You know that AC has gradient -2 and passes through point A, whose coordinates you already have as (-1, -2)
Angle ABC is the right angle meaning it is angle B not A so AB and BC are perpendicular
8. Angle C cuts the y axis meaning x=0 y+0=8 y=8

A= (0,-2)
C= (0,8)
D=(4,4)

AREA= BASE TIME HEIGHT/ 2

Base= 10
Height=4

40/2 = 20
9. (Original post by simonb451)
Angle C cuts the y axis meaning x=0 y+0=8 y=8

A= (0,-2)
C= (0,8)
D=(4,4)

AREA= BASE TIME HEIGHT/ 2

Base= 10
Height=4

40/2 = 20
thnxx a lot
10. (Original post by shana_t)
thnxx a lot
Thats fine- are you teaching yourself AS maths?
11. (Original post by simonb451)
Thats fine- are you teaching yourself AS maths?
yep, preparing for A level maths, gona start it from this september
12. (Original post by shana_t)
yep, preparing for A level maths, gona start it from this september
C1 and C2 are pretty good- Its S1 that most people seem to struggle with
13. (Original post by simonb451)
C1 and C2 are pretty good- Its S1 that most people seem to struggle with
ohh, my school does C1, C2 and D1. Wt about D1??
14. (Original post by shana_t)
ohh, my school does C1, C2 and D1. Wt about D1??
Oh we do D1 at A2, it sounds easy from what ive heard- the only tricky area of AS core was probably trig as it does take a while to get your head around sometimes
15. (Original post by simonb451)
Oh we do D1 at A2, it sounds easy from what ive heard- the only tricky area of AS core was probably trig as it does take a while to get your head around sometimes
oh okey
16. does anyone know how to work out this qs:
the straight line l passes through A(1, 3√3) and B (2+√3,3+4√3)
the gradient of l is √3
give the equation of l in the form of y=mx+c, where constants m and c are surds given in their simplest form.
17. You can use or rearrange to get .

Using the first option you will have which you will need to rearrange.

With the second option .
18. (Original post by BabyMaths)
You can use or rearrange to get .

Using the first option you will have which you will need to rearrange.

With the second option .
the answer is y=√3x+2√3, I still don't get it
19. (Original post by shana_t)
the answer is y=√3x+2√3, I still don't get it
You have two points You've being told that the gradient of the line l which passes through these two points is

Now the equation of a line is given by the formula 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 well from the question and for the coordinates of A correct?

Substitute these into the formula along with m, expand and rearrange into the form
20. (Original post by shana_t)
the answer is y=√3x+2√3, I still don't get it
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

You already know the equation is and you know that when .

You can use those values to find the value of c.

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