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Point of Intersection.

  1. Jan 24, 2008 #1
    The question asks that I : Find the point of intersection for each pair of lines.

    a) x + y= 4, x - 2y=1
    b) x + 2y= 0, x - y= 3
    c) 2x + y= 1, x + y= 2
    d) 6x= 12 - 3y, 1/2y - x= -5
    e) 1/2x - y=8, x + 1/3y= 2
    f) 5 + y= 4x, x + 2= 2/3y

    I understand the formula of y=mx+b, m being the slope and b being the y-intercept. But everytime I try to solve these problems, my answer is different than the one in the book.

    I figured I should change each of the equations from standard form to y=mx+b format.

    For example a)

    y=1/2x -1/2

    Am I doing something wrong? Because, when I try to draw it out on a graph my answer ends up being something weird like (2 1/3, 3/4).

    Also is there any other way I can pind the POI without drawing a graph?

    I need this for my exams...so please...help?
  2. jcsd
  3. Jan 24, 2008 #2


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    One suggestion for the graphical method in item (a) is use axes intercept feature of the equations since they are given in the standard form. What is x when y is zero; what is y when x is zero. Do this for both equations and draw each line.

    For your symbolic algebraic method, you took a good approach by find y of both equations and then equating them.

    (1/2)x-(1/2) = -x+4
    Results in (check for mistakes?) x=3, and then y=....
  4. Jan 24, 2008 #3
    I undestand now.

    Thank you very much for your help. ^___^
  5. Jan 24, 2008 #4


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    Of course there is. At the point, (x,y), where the two lines intersect, x and y must satisfy both equations. After you know, from the first equation, that y= -x+ 4, and, from the second that y= (1/2)x- 1/2, you know those two y values are the same thing: -x+ 4= (1/2)x- 1/2. Multiply both sides by 2: -2x+ 8= x- 1. Add 2x and 1 to both sides: 9= 3x so x= 9/3= 3. Then y= -3+ 4= 1. Notice that after you divided by 2 to find y in the second equation, I suggested that you immediately multiply by 2! After getting y= -x+ 4, you can just replace the "y" in x- 2y= 1 by that: x- 2(-x+ 4)= 1 so x+ 2x-8= 1. Now 3x-8= 1 so 3x= 9 again. As far as your graph is concerned, you should have used larger space. For example, after you saw that the intersection was around (2 1/3, 3/4), you might have drawn a new graph, just between, say x= 2 and x= 3, y= 0 and y= 1.

    Why would you need this for exams? If they never taught you to solve two equations except by graphing them, why not just do it that way?
  6. Jan 24, 2008 #5
    They did teach me. I just never really understood me. But I do now, so thanks for your troubles. :D
  7. Mar 31, 2010 #6
    Re: Point of Intersection/ Equation

    I am in Grade 9, and not neccesarily the most qualified to teach you anything, but I created a simpler equation a few days ago in math class than making the equations equal.
    First, you are going to solve for the x-coordinate of the P.O.I.
    To do this, use this formula: x= "d" divided by "S", with d being equal to B1 - B 2. "S" is equal to the difference in slope, using M2 subtract M1. This should give you the x coordinate for your P.O.I.

    ex. y= 8x+18 d would equal 18, (18-0)
    y= 10x S would equal 2, (10-8)
    x= d divided by S
    = 18 divided by two
    = 9

    Therefore, your x value would be 9. Then you just substitute in the x value in either equation, then calculate to find P.O.I.

    ex. y= 10x
    = 10(9)
    = 90 Your coordinate would be (10, 90)

    Checking with other equation:

    y`= 8x + 18
    = 8(9) + 18
    = 72+ 18
    = 90 Again, the y value comes out as 90, so the POI must be 10, 90

    I know this is somewhat outdated, by the way.
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