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Simple gradient/graph question.

  1. Apr 23, 2008 #1
    1. The points A and b have Coords (6,-1) and (2,5) respectively

    a.Show gradient is - 3/2

    -1-5 = -6
    6-2 = 4

    = -3/2

    Hence find the equation of the line AB with answer in form Ax + By = C

    A B C are integers.

    - now this isn't a homework question im just doing some extra work in my own time. I have never been taught this, but when looking through past papers i have seen that it has actualy come up, can someone explain to me how exactly you do these. I have attempted it, and know the answer because i have the mark scheme, but i don't 100% understand it.

    i know you use y-5 and y+1 somewhere, which is the inverse of the Y coordinates in the 2 points im given. and i know that you somehow times the X by the gradient with those coordinates inverted but im not sure how it works.

    answer = 3x+2y = 16 can anyone explain this?
  2. jcsd
  3. Apr 23, 2008 #2


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    Staff Emeritus
    Science Advisor

    You got the gradient by (1) subtracting the two y values, (2) subtracting the two x values, (3) dividing the first by the second.

    And, of course, you could do that for any two points on the line and get the same thing. Suppose you pick an arbitrary point on the line and call it (x, y). Now use (x,y) and (2, 5) to calculate the gradient. You would have (y- 5)/(x- 2) and that must give you the same thing as (-1-5)/(6-2)= -3/2. That is, you must have (y- 5)/(x-2)= -3/2. "Cross multiplying", that is, multiplying both sides by 2 and (x-2), we get 2(y- 5)= -3(x- 2). Multiply those out, move the "x" term to the left, and the constants to the right and see what you get.
  4. Apr 23, 2008 #3


    next question is to find equation of perpendicular line that crosses through point b.

    is that (y-5)/(X-2) = 2/3

    =3y-15 = 2x-4
    or is something wrong
    Last edited: Apr 23, 2008
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