Question

Find the equation of a line that passes through (3,-1) and is perpendicular to the equation 3x+y=2

Answer 1

y = (1/3)x - 2

Step-by-step explanation:

First, we need to identify the slope of the equation 3x + y = 2, so we need to solve for x as:

`\begin{gathered} 3x+y=2 \\ y=2-3x \end{gathered}`

Now, the number beside the variable x is the slope, so the slope is -3

Then, two lines are perpendicular if the product of both slopes is equal to -1. It means that the slope m for our equation should be:

`\begin{gathered} -3\cdot m=-1 \\ m=(-1)/(-3)=(1)/(3) \end{gathered}`

Now, we can find the equation of the line using the following:

`y=m(x-x_1)+y_1`

Where m is the slope and (x1, y1) is a point on the line. So, replacing my 1/3 and (x1, y1) by (3, -1), we get:

`\begin{gathered} y=(1)/(3)(x-3)+(-1) \\ y=(1)/(3)x-(1)/(3)\cdot3-1 \\ y=(1)/(3)x-1-1 \\ y=(1)/(3)x-2 \end{gathered}`

Therefore, the equation of the line is: y = (1/3)x - 2