Question

Find the equation of the line through (-8,1) which is perpendicular to the line y= -x/2-6

Answer 1

Let's start with the standard form of a line:

`y=mx+b`

"m" is the slope of this line.

The slopes of two perpendicular lines, m1 and m2, have the relation:

`m_2=-(1)/(m_1)`

So, if we want the slope of a line perpendicular to a known line, we can simply use its slope to it.

The given line here is:

`\begin{gathered} y=-(x)/(2)-6 \\ y=-(1)/(2)x-6 \end{gathered}`

So, the slope of this line is:

`m_1=-(1)/(2)`

To find the slope of a lie perpendicular to it, we use the above relation:

`m_2=-(1)/(-(1)/(2))=-((1)/(1))/(-(1)/(2))=-(1)/(1)\cdot(-(2)/(1))=2`

So, the slope of our line is 2.

In the slope-intercept form, this is:

`\begin{gathered} y-y_0=m_2(x-x_0) \\ y-y_0=2(x-x_0) \end{gathered}`

Now, we just need to have x0 and y0, which are the coordinates of a point in the line. We are given the point (-8,1), which means that:

`\begin{gathered} x_0=-8 \\ y_0=1 \end{gathered}`

So:

`\begin{gathered} y-y_0=2(x-x_0)_{} \\ y-1=2(x-(-8)) \\ y-1=2(x+8) \\ y-1=2x+16 \\ y=2x+16+1 \\ y=2x+17 \end{gathered}`