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# Co-ordinate Geometry Help watch

1. Hi there,

1) Find the equation of the line joining the points (2,-3) and (5,12)

2) Find the co-ordinates of the point of intersection of the lines with equations
3x+y=1 and y=7x-2

I'm not sure how to start here.

3) Find the equation of the circle with centre (3,4) which passes through the point (8,-8)

4) Show that the line y=3x-7 is a tangent to the curve y=2x^2-5x+1
Find the co-ordinates of the point of contact.
I've done
3x-7=2x^2-5x+1
3x=2x^2-5x+8
y=2x^2-8x+8

Is this the correct method?

-------

Again, thanks a lot. I do have more questions but I figured it'd be too much in one go.

2. To see if your equations are right, you just need to substitute the points back in.

Let's start at question 1:

y = 5x + 12 and point (2, -3) - does this work?
3. (Original post by Mr M)
To see if your equations are right, you just need to substitute the points back in.

Let's start at question 1:

y = 5x + 12 and point (2, -3) - does this work?
-_- Fail by me. Didn't work out C.

5x-13?
4. (Original post by tehforum)
-_- Fail by me. Didn't work out C.

5x-13?
That one is right now. Second one is solving a pair of simultaneous equations.
5. For number 1 :

For number 2 :
3x+y=1 and y=7x-2

Rearrange the first to get it in the form y=, then make them equal each other / move to one side / factorise to get x . Then plug in to get Y.

Havent done 3 yet .
6. (Original post by Mr M)
That one is right now. Second one is solving a pair of simultaneous equations.
Second one.

3x+y=1
y=7x-2

Sub 2 into 1

3x+7x-2=1
10x-2=1
10x=3
x=0.3

Sub then x=0.3 into 1
3(0.3)+y=1
0.9+y=1
y=0.1
7. (Original post by tehforum)
Second one.

3x+y=1
y=7x-2

Sub 2 into 1

3x+7x-2=1
10x-2=1
10x=3
x=0.3

Sub then x=0.3 into 1
3(0.3)+y=1
0.9+y=1
y=0.1
Yes, third looks correct to me too.

You started 4 correctly but then did something weird. Form a quadratic in x and work out the discriminant - what will that tell you that is relevant?
8. 3x-7=2x^2-5x+1

This is correct? I know you've given me instructions, but I think I may need a further budge.

Re: discriminant is it needed? b^2-4ac > 0 = 2 real roots.
9. (Original post by tehforum)
3x-7=2x^2-5x+1

This is correct? I know you've given me instructions, but I think I may need a further budge.

Re: discriminant is it needed? b^2-4ac > 0 = 2 real roots.

The further nudge is you don't want to find that the discriminant has 2 real roots. What do you know about tangents to curves?
10. (Original post by Mr M)

The further nudge is you don't want to find that the discriminant has 2 real roots. What do you know about tangents to curves?
3x-7=2x^2-5x+1
3x=2x^2-5x+8
y=2x^2-8x+8
y=x^2-4x+4
y= (x-2)^2

So x= 2.

When sub that into 3x-7
6-7 = -1

x =2, y=-1

(2,-1)
11. (Original post by tehforum)
3x-7=2x^2-5x+1
3x=2x^2-5x+8
y=2x^2-8x+8
y=x^2-4x+4
y= (x-2)^2

So x= 2.

When sub that into 3x-7
6-7 = -1

x =2, y=-1

(2,-1)
Why have you reintroduced y in the first section?

Why are you substituting? Were you asked to find a pair of coordinates?

Visualise what is going on.

You have a positive parabola so it is U shaped.

You have a straight line with a positive gradient like this /

If it is a tangent there should be one point of intersection.

Trying to solve a pair of simultaneous equations and showing there is only one solution will show it is a tangent. Alternatively, you can form a quadratic and set it equal to zero (not y). If you find the discriminant you should get one real root to show there is only one answer.

Edit: Whoops yes you were! Your coordinates are OK. I still don't like the unnecessary y and you need to write a sentence to explain why you know you have found a tangent.

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