Question

What is the equation of the line that contains the points ( -1 , 1 ) and ( 5 , -3) ?

Answer 1

`y=-(2)/(3)x+(1)/(3)`

Explanation:

The equation of the line in its slope and intercept form is as follows:

`\begin{gathered} y=mx+b \\ \text{ where m is the slope and b is y-intercept} \end{gathered}`

The slope can be calculated as follows:

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

Replacing the points (-1 , 1) and (5 , -3) :

`m=(-3-1)/(5-(-1))=(-4)/(5+1)=-(4)/(6)=-(2)/(3)`

Now, we calculate the value of b, with the help of the slope and the point (-1,1)

`\begin{gathered} 1=-(2)/(3)\cdot-1+b \\ b=1-(2)/(3) \\ b=(1)/(3) \end{gathered}`

Therefore, the equation would be:

`y=-(2)/(3)x+(1)/(3)`