Hey there! Sign in to join this conversationNew here? Join for free

Maths C1 Coordinate Geometry Watch

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
    • Very Important Poster
    Offline

    21
    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?
    • Community Assistant
    • Welcome Squad
    Offline

    19
    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 :/
    • Very Important Poster
    Offline

    21
    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)
    • Very Important Poster
    Offline

    21
    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
    • Community Assistant
    • Welcome Squad
    Offline

    19
    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?!!?!?!
    • Community Assistant
    • Welcome Squad
    Offline

    19
    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
    • Very Important Poster
    Offline

    21
    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
    • Community Assistant
    • Welcome Squad
    Offline

    19
    ReputationRep:
    (Original post by seanfm)
    correct answer but minus swag points for not using gaussian elimination. Sorry
    this is why i'll fail c1
    • Community Assistant
    • Welcome Squad
    Offline

    19
    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:
    • Very Important Poster
    Offline

    21
    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

    13
    ReputationRep:
    solve the simultaneous equations then BAM you get x coordinate and y coordinate
 
 
 
  • 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.

  • Poll
    Break up or unrequited love?
    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
  • 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.

  • 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.