Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    12
    ReputationRep:
    Name:  Capture.PNG
Views: 37
Size:  10.1 KB

    Anyone know how to solve this STEP BY STEP!
    Offline

    17
    ReputationRep:
    (Original post by NothingButWaleed;[url="tel:67109102")
    67109102[/url]]Name:  Capture.PNG
Views: 37
Size:  10.1 KB

    Anyone know how to solve this STEP BY STEP!
    1) Work out the gradient AB
    2) find M
    3) substitute the point of c into y1-y=m(x-x1)

    Y1 = the y coordinate of C
    X1 = the X coordinate of X

    X and Y stay the same so you just substitute numbers in.
    Offline

    12
    ReputationRep:
    Okay, so the coordinates of A and B are (-3, 0) and (0,5) respectively.
    To find the gradient of line AB, you just find the difference in y and divide by the difference in x. 0-5/-3-0 = -5/-3 = 5/3, so the line AB has an equation like this: y =5/3x + c.

    We're told to find the perpendicular of AB, which is just the negative reciprocal of the gradient. So, you basically flip and add a negative sign 5/3 -> -3/5. So the equation of the line we want is y = -3/5x + c.

    It also passes through C which is (4, -2), so we put that into the equation y = -3/5x+c.
    -2 = -3/5(4) + c
    -2 = -12/5 + c
    -2 + 12/5 = c
    c = 2/5

    Therefore, the equation of the line which is perpendicular to AB and passes through C is: y = -3/5x + 2/5. Hope I've explained it clearly enough.


    Posted from TSR Mobile
    • Community Assistant
    • Welcome Squad
    Online

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by Aamna-)
    Okay, so the coordinates of A and B are (-3, 0) and (0,5) respectively.
    To find the gradient of line AB, you just find the difference in y and divide by the difference in x. 0-5/-3-0 = -5/-3 = 5/3, so the line AB has an equation like this: y =5/3x + c.

    We're told to find the perpendicular of AB, which is just the negative reciprocal of the gradient. So, you basically flip and add a negative sign 5/3 -> -3/5. So the equation of the line we want is y = -3/5x + c.

    It also passes through C which is (4, -2), so we put that into the equation y = -3/5x+c.
    -2 = -3/5(4) + c
    -2 = -12/5 + c
    -2 + 12/5 = c
    c = 2/5

    Therefore, the equation of the line which is perpendicular to AB and passes through C is: y = -3/5x + 2/5. Hope I've explained it clearly enough.


    Posted from TSR Mobile
    Let's restrain from posting full solutions, shall we? We can only point him/her in the right direction otherwise we'd be doing his/her maths for him/her without him/her putting thought into it. The post above is sufficient help.
 
 
 
  • 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
    Has a teacher ever helped you cheat?
    Useful resources

    Study tools

    Rosette

    Essay expert

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

    Thinking about uni already?

    Thinking about uni already?

    See where you can apply with our uni match tool

    Student chat

    Ask a question

    Chat to other GCSE students and get your study questions answered.

    Creating

    Make study resources

    Create all the resources you need to get the grades.

    Planner

    Create your own Study Plan

    Organise all your homework and exams so you never miss another deadline.

    Resources by subject

    From flashcards to mind maps; there's everything you need for all of your GCSE subjects.

    Papers

    Find past papers

    100s of GCSE past papers for all your subjects at your fingertips.

    Help out other students

    Can you help? Study help unanswered threads

    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

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