Custom Search
If you want to discuss more on this or other issues related to Physics, feel free to leave a message on my Facebook page.

Sunday, November 14, 2010

y = mx + c

We have seen in a previous post about the gradient and how it is calculated. We have also learned about the y-intercept. We are now going to combined the two concept to learn about the equation of a straight line.

All straight line can be represented by a equation that will be in the form of 

y = mx + c

We are now going to use a straight line in order to see how the equation above can be determine and how the x-coordinate or the y-coordinate could be determined if one of the two and the equation is available.


Fig 1

Fig 1 above shows a straight line graph.

The gradient of the graph can be determined using the method described here.

Two coordinates that can be used are (1,6) and  (5,18)

Thus Gradient = (18 –6)/(5-1)

                               = (12)/(4)

                                 = 3

And using the methods described here we can deduce that the y-intercept is 3.

Now the gradient = 3

and y-intercept =3

If we say that gradient = m then m= 3

and if we say that y-intercept = c then c=3

replacing the values for m and c in y= mx +c

the equation of the line becomes

y = 3x + 3

All straight lines can be represented with an equation and the way to obtain the equation is by using the method above.

We are going to use the equation to determine

(i) the value of the y-coordinate when the x-coordinate is 3

(ii) the value of the x coordinate when the y-coordinate is 15

(i) Equation of the line is y = 3x + 3 and x=3

replacing x=3 in the equation we get

y =3(3) +3


=12 check on the graph!!!!!!

(ii) y =15

Replacing into the equation we get

15 = 3x +3

15 – 3 = 3x

12 = 3x

x =12/3

x = 4 Check on the graph!!!!!!


Related Posts Plugin for WordPress, Blogger...