Results 1 to 2 of 2

Thread: equation of a line

  1. #1
    Senior Member euclid2's Avatar
    Joined
    May 2008
    From
    Ottawa, Canada
    Posts
    400
    Awards
    1

    equation of a line

    Determine the equation of a line with an x-intercept of 4 and passing through the point(5,2)

    Am I doing it correctly?
    y = m(x - p) + q
    y=4(x-5)+2
    y=4x-20+2
    y=4x-18

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by euclid2 View Post
    Determine the equation of a line with an x-intercept of 4 and passing through the point(5,2)

    Am I doing it correctly?
    y = m(x - p) + q
    y=4(x-5)+2
    y=4x-20+2
    y=4x-18

    Thanks
    Explanation:
    You know 2 points which are $\displaystyle (4, 0)$ [as you are told it passes x-axis through $\displaystyle x=4$ which occurs when $\displaystyle y=0$] and $\displaystyle (5,2)$.

    An equation of a straight line is in the form: $\displaystyle y=mx+c$ where $\displaystyle x$ and $\displaystyle y$ are your variables, $\displaystyle m$ is your gradient and $\displaystyle c$ is your y-intercept.

    Gradient is calculated by $\displaystyle m = \frac{y_1-y_2}{x_1 - x_2}$ where $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$ are two co-ordinates.

    Use this gradient value and any co-ordinates to find the general form of the equation using $\displaystyle y-y_1 = m (x - x_1)$ where $\displaystyle (x_1, y_1)$ are any co-ordinates of the line.


    The method and working out:
    The two co-ordinates are $\displaystyle (4, 0)$ and $\displaystyle (5,2)$.

    $\displaystyle m = \frac{y_1-y_2}{x_1 - x_2} = \frac{2-0}{5-4} = \frac21 = 2$

    $\displaystyle y-y_1 = m (x - x_1) \implies y-0 = 2(x-4) \implies y = 2x-8$

    Check this with the other co-ordinates, $\displaystyle (5,2)$.

    $\displaystyle y = 2x-8 \implies 5 = 2(2) - 8 $

    Therefore the equation of the line is $\displaystyle y=2x-8$.
    Last edited by Simplicity; Jul 10th 2008 at 10:29 AM. Reason: Grammer, Spelling ^_^
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Mar 18th 2011, 10:36 AM
  2. Replies: 9
    Last Post: Jul 31st 2009, 05:39 AM
  3. Replies: 11
    Last Post: Jun 2nd 2009, 06:08 PM
  4. Replies: 5
    Last Post: Oct 13th 2008, 10:16 AM
  5. equation of a line perpendicular to the line defined
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Jun 6th 2007, 09:02 AM

Search Tags


/mathhelpforum @mathhelpforum