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:

$m_2=-\frac{1}{m_1}$

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:

$\begin{gathered} y=-\frac{x}{2}-6 \\ y=-\frac{1}{2}x-6 \end{gathered}$

So, the slope of this line is:

$m_1=-\frac{1}{2}$

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

$m_2=-\frac{1}{-\frac{1}{2}}=-\frac{\frac{1}{1}}{-\frac{1}{2}}=-\frac{1}{1}\cdot(-\frac{2}{1})=2$

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:

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