Maths C1 Coordinate Geometry

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    I'm stuck on a question that I remember doing but I've ended up forgetting.

    Coordinates given for ABCD (-5,0),(0,5),(3,4)(4,-3)

    Lines AC and BD intersect at E.
    > Show that the coordinates of E are (1,3)

    How do I work out the intersection point?

    Notes: AC Gradient = 1/2
    BD Gradient = -2/1
    (They're perpendicular)

    Equation AC is x-2y+5 = 0
    Equation BD is y=-2x+5

    Please help
    Offline

    3
    ReputationRep:
    (Original post by slot_)
    I'm stuck on a question that I remember doing but I've ended up forgetting.

    Coordinates given for ABCD (-5,0),(0,5),(3,4)(4,-3)

    Lines AC and BD intersect at E.
    > Show that the coordinates of E are (1,3)

    How do I work out the intersection point?

    Notes: AC Gradient = 1/2
    BD Gradient = -2/1
    (They're perpendicular)

    Equation AC is x-2y+5 = 0
    Equation BD is y=-2x+5

    Please help
    If you draw two lines on a graph and make them intersect, you will see that their intersection has the same co-ordinates, which means the same point is on the two different lines and so satisfies both equations.

    Does that give you a hint as to how you can solve those two equations at the same time?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by SeanFM)
    If you draw two lines on a graph and make them intersect, you will see that their intersection has the same co-ordinates.

    Does that give you a hint as to how you can solve those two equations at the same time?
    Simultaneous?
    Offline

    3
    ReputationRep:
    (Original post by SeanFM)
    If you draw two lines on a graph and make them intersect, you will see that their intersection has the same co-ordinates.

    Does that give you a hint as to how you can solve those two equations at the same time?
    Hmmmmmmmmmmmmmmmmmmm..........

    Same time, eh?........


    Yeh I don't get it :/
    Offline

    3
    ReputationRep:
    (Original post by slot_)
    Simultaneous?
    Precisely
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by SeanFM)
    Precisely
    Derp, I got it now. Just needed the pointer lol

    For anyone who may want to correct me

    AC Gradient = 1/2, Equation = 2y = x+5.
    BD = 2y = -2x + 5 with Gradient of -2

    -2x + 5 = 1/2x + 5/2
    when you multiply by 2 to get whole numbers
    -4x + 10 = x + 5
    rearrange
    (-5) + 10 = x + 4x
    5 = 5x
    x = 1
    Substitute in to later equation so 2y = x + 5
    2y = 1 + 5
    2y = 6
    y = 6/3
    y = 3

    (1,3)
    Offline

    3
    ReputationRep:
    (Original post by slot_)
    Derp, I got it now. Just needed the pointer lol

    For anyone who may want to correct me

    AC Gradient = 1/2, Equation = 2y = x+5.
    BD = 2y = -2x + 5 with Gradient of -2

    -2x + 5 = 1/2x + 5/2
    when you multiply by 2 to get whole numbers
    -4x + 10 = x + 5
    rearrange
    (-5) + 10 = x + 4x
    5 = 5x
    x = 1
    Substitute in to later equation so 2y = x + 5
    2y = 1 + 5
    2y = 6
    y = 6/3
    y = 3

    (1,3)
    A good way to check your answer is to see if the point satisfies both equations.. after all, that is what you're finding
    Offline

    3
    ReputationRep:
    (Original post by SeanFM)
    Precisely
    OKAY I THINK I GOT IT

    Step 1: Rearrange equations
    x-2y=-5
    2x+y=5

    Step 2: Represent in matrix form

    \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}  = \begin{bmatrix} -5 \\ 5 \end{bmatrix}

    Step 3: Find the inverse matrix

    \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}^{-1} = \frac{1}{5}\begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}

    Step 4: Pre-multiply by the inverse matrix
    \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}^{-1} \begin{bmatrix} -5 \\ 5 \end{bmatrix}

    \Rightarrow \mathbb{I} \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} -5 \\ 5 \end{bmatrix}

    Step 5: Clean up
    \begin{bmatrix} x \\ y \end{bmatrix}  = \frac{1}{5}\begin{bmatrix} 5 \\ 15 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}

    So:

    x=1
    y=3


    DID I GET IT RIGHT?!!?!?!
    Offline

    3
    ReputationRep:
    (Original post by slot_)
    Derp, I got it now. Just needed the pointer lol

    For anyone who may want to correct me

    AC Gradient = 1/2, Equation = 2y = x+5.
    BD = 2y = -2x + 5 with Gradient of -2

    -2x + 5 = 1/2x + 5/2
    when you multiply by 2 to get whole numbers
    -4x + 10 = x + 5
    rearrange
    (-5) + 10 = x + 4x
    5 = 5x
    x = 1
    Substitute in to later equation so 2y = x + 5
    2y = 1 + 5
    2y = 6
    y = 6/3
    y = 3

    (1,3)
    ur methud is confusing :s
    Offline

    3
    ReputationRep:
    (Original post by RDKGames)
    OKAY I THINK I GOT IT

    Step 1: Rearrange equations
    x-2y=-5
    2x+y=5

    Step 2: Represent in matrix form

    \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}  = \begin{bmatrix} -5 \\ 5 \end{bmatrix}

    Step 3: Find the inverse matrix

    \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}^{-1} = \frac{1}{5}\begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}

    Step 4: Pre-multiply by the inverse matrix
    \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}^{-1} \begin{bmatrix} -5 \\ 5 \end{bmatrix}

    \Rightarrow \mathbb{I} \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} -5 \\ 5 \end{bmatrix}

    Step 5: Clean up
    \begin{bmatrix} x \\ y \end{bmatrix}  = \frac{1}{5}\begin{bmatrix} 5 \\ 15 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}

    So:

    x=1
    y=3


    DID I GET IT RIGHT?!!?!?!
    Correct answer but minus swag points for not using gaussian elimination. Sorry
    Offline

    3
    ReputationRep:
    (Original post by seanfm)
    correct answer but minus swag points for not using gaussian elimination. Sorry
    this is why i'll fail c1
    Offline

    3
    ReputationRep:
    (Original post by SeanFM)
    Correct answer but minus swag points for not using gaussian elimination. Sorry
    On a serious note, do you reckon they would actually award marks for this method in C1??? :hmmmm:
    Offline

    3
    ReputationRep:
    (Original post by RDKGames)
    On a serious note, do you reckon they would actually award marks for this method in C1??? :hmmmm:
    I'd hope so. May require some consultation with a senior exam officer or something, and some investigation into how and why the student is using such a method. Fancy but time consuming and superfluous
    Offline

    2
    ReputationRep:
    solve the simultaneous equations then BAM you get x coordinate and y coordinate
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: September 28, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
What is your favourite day of the week
Help with your A-levels

All the essentials

The adventure begins mug

Student life: what to expect

What it's really like going to uni

Rosette

Essay expert

Learn to write like a pro with our ultimate essay guide.

Uni match

Uni match

Our tool will help you find the perfect course for you

Study planner

Create a study plan

Get your head around what you need to do and when with the study planner tool.

Study planner

Resources by subject

Everything from mind maps to class notes.

Hands typing

Degrees without fees

Discover more about degree-level apprenticeships.

A student doing homework

Study tips from A* students

Students who got top grades in their A-levels share their secrets

Study help links and info

Can you help? Study help unanswered threadsRules and posting guidelines

Sponsored content:

HEAR

HEAR

Find out how a Higher Education Achievement Report can help you prove your achievements.

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.