What is the equation of the line passing through the points #(4/5, 1/5)# and #(1/2, 3/2)#?

1 Answer
Aug 24, 2017

#y=-4.33x+3.665#

Explanation:

Since #y=mx+c#, m being the gradient and c being the intercept, we have to format the equation like this.

The calculations will be easier if we convert this into decimals, so we'll do just that:
#4/5=0.8#

#1/5=0.2#

#1/2=0.5#

#3/2=1.5#

From then on, we'll calculate the gradient using the equation:

#m=(y_2-y_1)/(x_2-x_1)#

#=(1.5-0.2)/(0.5-0.8)#

Notice how I put a set of coordinates to one side and another set of coordinates to another. This should give you:

#1.3/-0.3#

#=-4.33#

Now, we substitute this into the equation #y=mx+c# to get:

#y=-4.33x+c#

From here, its all a matter of substituting things in. You input a y value from one of your set of coordinates and the corresponding x value like so:

#1.5 = -4.33(0.5)+c#

So, in order to get c, you need to 'transfer' the gradient and the x value to the left hand side and 'eliminate' those variables from the right hand side:

#1.5 + 4.33(0.5)= -4.33(0.5)+c+4.33(0.5)#

#1.5 + 4.33(0.5) = c

#c=3.665#

So now, the equation you get is:

#y=-4.33x+3.665#

You can check if these values are correct by substituting the letters for your other set of coordinates.

Hope this helps!