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# C1- Coordinate Geometry watch

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1. Dear Users,

I found it hard to find the following question.

Question1
The line y = x and y = 2x -5 intersect at the point A. Find the equation of the line with gradient 2/5 that passes trough the point A. (Hint: Solve y =x and y = 2x -5)
I might be able to work it out if I can solve them simultaniously.

Question 2
The lines y=4x-10 and y = x-1 intersect at the point T. Find the equation of the line with gradient -2/3 that passes trhough the point T. Write your answer to the form ax + by + x= 0, where a, b and c are integers.
2. (Original post by Limelight31)
Dear Users,

I found it hard to find the following question.

Question1
The line y = x and y = 2x -5 intersect at the point A. Find the equation of the line with gradient 2/5 that passes trough the point A. (Hint: Solve y =x and y = 2x -5)
I might be able to work it out if I can solve them simultaniously.

Question 2
The lines y=4x-10 and y = x-1 intersect at the point T. Find the equation of the line with gradient -2/3 that passes trhough the point T. Write your answer to the form ax + by + x= 0, where a, b and c are integers.
What you've said for Q1 is right - so how can you solve those two equations simultaneously, and what do you do with the answer?
3. Q2

I did this;
>4x -10 = x -1
>4x-x=-1+10
>3x=9
x=3
Intercept being of lines is on (3,0)
And thus y-y1=m(x-x1)
And thus y-0=-2/3(x-3)
>y=-2/3x+2
multiply by 3 for removing fraction
3y=-2x+6
2x+3y-12=0
4. (Original post by SeanFM)
What you've said for Q1 is right - so how can you solve those two equations simultaneously, and what do you do with the answer?
That is what the book states, that I should simultaniously solve it.

Not sure where to start as it has no real values.
5. (Original post by Limelight31)
That is what the book states, that I should simultaniously solve it.

Not sure where to start as it has no real values.
Okay. Do you know why you want to solve those two equations?

And if I said the word substitution, would that help you solve it?
6. (Original post by Limelight31)
Q2

I did this;
>4x -10 = x -1
>4x-x=-1+10
>3x=9
x=3
Intercept being of lines is on (3,0)
And thus y-y1=m(x-x1)
And thus y-0=-2/3(x-3)
>y=-2/3x+2
multiply by 3 for removing fraction
3y=-2x+6
2x+3y-12=0
Good job so far

It was going really well until after the line x=3 - you got y= 0, but how?
7. I could substitute it maybe, This is my working out for that :
y=x , y= 2x - 5
___
putting all x on one side;(used book info_)
x=y
2x=-y-5
x=-1/2y-5/2y
8. (Original post by SeanFM)
Good job so far

It was going really well until after the line x=3 - you got y= 0, but how?
haha genious so y - 1 would be 0 and y-0 stays y, or would that be 0 aswell "D
9. (Original post by Limelight31)
haha genious so y - 1 would be 0 and y-0 stays y, or would that be 0 aswell "D
You're looking for the point T, which will help you answer the question. You have found that the x coordinate of T is 3. Both of the equations are satisfied by the coordinates of T (because T is on both lines). So what is the y coordinate of T?
10. (Original post by SeanFM)
You're looking for the point T, which will help you answer the question. You have found that the x coordinate of T is 3. Both of the equations are satisfied by the coordinates of T (because T is on both lines). So what is the y coordinate of T?
so the point of intercept is not 3,0 but would be putting in the x on one of the equations, and thus getting (3,2) mate?
11. (Original post by Limelight31)
so the point of intercept is not 3,0 but would be putting in the x on one of the equations, and thus getting (3,2) mate?
Correct - well done

What is the next step to answer the question?
12. haha thanks mate, the rest is putting in the values in h formula, and it leads to answer, im sttill having some struggle with the Q1.

Substitution is the word u suggested, after looking up in the book (few chapters back) i came to the idea of bringing all x'és or y'es
to one site and start from there.
Let me work it out:
Y= x
Y = 2x -5
using the substitution
x=y
2x=-y-5
x=-1/2y-5/2

Then substituting right? I dont know what to do man

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