Sec 2 Maths

• BillaMax

  10 Jan 10, 17:37

  

  How to solve the following question

  Question 1 will do

  

• JCMaths

  10 Jan 10, 18:07

  For linear equation, y = mx + c, m is the gradient while c is the y-intercept.

  Look in your textbook for the formula for gradient using 2 points (x1,y1) and (x2,y2). You can pick any 2 points on the graph given to use.

  For the value of c, once you have m, you can find c using any pair of (x,y) values (or actually it's obvious from the graph even without calculation, but since the question says 'find' and not 'state', I personally prefer to show some working).

• BillaMax

  10 Jan 10, 18:35

  Originally posted by JCMaths:

  For linear equation, y = mx + c, m is the gradient while c is the y-intercept.

  Look in your textbook for the formula for gradient using 2 points (x1,y1) and (x2,y2). You can pick any 2 points on the graph given to use.

  For the value of c, once you have m, you can find c using any pair of (x,y) values (or actually it's obvious from the graph even without calculation, but since the question says 'find' and not 'state', I personally prefer to show some working).

  so how do i do the question?

• qdtimes2

  10 Jan 10, 19:00

  i will use the 2 end points of the line, (4, 5) and (12, -5).

  1.gradient, m = (-5- 5)/(12- 4)

                       = -1.25

  2.i will use one of the end point again.

      y = mx + c

      5 = -1.25(4) + c

       c = 10

• JCMaths

  10 Jan 10, 19:43

  Hmm.. Personally, I would prefer to refrain from giving out answers directly because I don't think it helps in the learning, if it can be just copied and submitted, without any thought process or effort put in. Has TS looked up the formula for gradient?