Question

Write an equation in general form for the line passing through (-12,-1) and perpendicular to the line whose equation is 6x-y-4=0

Answer 1

The Solution:

Given the equation of a line:

$6x-y-4=0$

We are required to find the equation of a line that is perpendicular to the given line and passes through the point (-12,-1).

Step 1:

Determine the slope of 6x-y-4=0.

$\begin{gathered} \text{ Express y in terms of x.} \\ 6x-4=y \\ y=6x-4 \end{gathered}$

The coefficient x, after making y the subject of the formula, is the slope of the given line. That is:

$\begin{gathered} \text{ The slope of }6x-y-4=0\text{ is 6} \\ \\ Slope=m_1=6 \end{gathered}$

Step 2:

Find the slope of the required line:

Since the two lines are perpendicular, the slope is:

$\begin{gathered} m_2=\frac{-1}{m_1}=\frac{-1}{6} \\ \end{gathered}$

Step 3:

Find the equation of the line that passes through point (-12,-1)

By formula,

$y-y_1=m_2(x-x_1)$

In this case,

$\begin{gathered} x_1=-12 \\ y_1=-1 \end{gathered}$

Substituting these values in the formula, we get the required equation is:

$\begin{gathered} y--1=\frac{-1}{6}(x--12) \\ \\ y+1=-\frac{1}{6}(x+12) \end{gathered}$

Cross multiplying, we get

$\begin{gathered} 6y+6=-x-12 \\ \\ 6y+x+6+12=0 \\ \\ 6y+x+18=0 \end{gathered}$

Therefore, the correct answer is:

$6y+x+18=0$