Question

what is the equation of a line that is perpendicular to the line shown and goes through the point (3,-1)

Answer 1

For lines to be perpendicular, the slope of one of them must be the inverse negative of the other one. So for example given the lines:

`y=mx+b`
`y=nx+c`

For them to be perpendicular their slopes must be:

`n=-(1)/(m)`

To determine the equation of a line perpendicular to the given one, the first step is to calculate the slope of the graphed line.

Using two points of the line, for example: (0,3) and (4,6)

`\begin{gathered} m=(y_1-y_2)/(x_1-x_2) \\ m=(6-3)/(4-0)=(3)/(4) \end{gathered}`

The slope of the graphed line is

`m=(3)/(4)`

Now the slope of the perpendicular line must be the inverse negative so that:

`\begin{gathered} n=-(1)/(m) \\ n=-(4)/(3) \end{gathered}`

Now that we know the slope of the perpendicular line, and is given that it crosses the point (3,-1) we can use the point slope form to determine its equation

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

For our line

`\begin{gathered} y-(-1)=-(4)/(3)(x-3) \\ y+1=-(4)/(3)x+4 \\ y=-(4)/(3)x+4-1 \\ y=-(4)/(3)x+3 \end{gathered}`

The equation of a perpendicular line of the one shown in the graph rthat goes through point (3,-1) is

`y=-(4)/(3)x+3`