Question

Line a has an equation of y = -1/3 x + 6. Line b includes the point (-1, 2) and isparallel to line a. What is the equation of line b?

Answer 1

hello

the question here request we find the equation of line passing through point (-1, 2)

the first equation given is

`y=-(1)/(3)x+6`

passing through points (-1, 2)

equation of line is given as

`\begin{gathered} y-y_1=m(x-x_1) \\ m=-(1)/(3) \\ \end{gathered}`

the points are (-1, 2)

`\begin{gathered} x_1=-1 \\ y_1=2 \end{gathered}`

now, let's substitute the values into the equation above

`\begin{gathered} y-y_1=m(x-x_1) \\ y-2=-(1)/(3)(x-(-1)) \\ y-2=-(1)/(3)(x+1) \\ y-2=-((x+1))/(3) \\ \text{cross multiply both sides} \\ 3(y-2)=-x-1 \\ 3y-6=x-1 \\ 3y=-x-1 \\ \text{divide both sides by the coeffiecient of y which is 3} \\ (3y)/(3)=(-x-1)/(3) \\ y=(-x-1)/(3) \\ y=-(x)/(3)-(1)/(3) \end{gathered}`