Answer:
An equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:
Step-by-step explanation:
We know that the slope-intercept form of the line equation is
y=mx+b
where m is the slope and b is the y-intercept.
Given the line
6x+y=2
Simplifying the equation to write into the slope-intercept form
y = -6x+2
So, the slope = -6
As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line.
Thus, the slope of the perpendicular line will be: -1/-6 = 1/6
Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be
![y-y_1=m\left(x-x_1\right)](https://tex.z-dn.net/?f=y-y_1%3Dm%5Cleft%28x-x_1%5Cright%29)
substituting the values m = 1/6 and the point (6, -2)
![y-\left(-2\right)=\frac{1}{6}\left(x-6\right)](https://tex.z-dn.net/?f=y-%5Cleft%28-2%5Cright%29%3D%5Cfrac%7B1%7D%7B6%7D%5Cleft%28x-6%5Cright%29)
![y+2=\frac{1}{6}\left(x-6\right)](https://tex.z-dn.net/?f=y%2B2%3D%5Cfrac%7B1%7D%7B6%7D%5Cleft%28x-6%5Cright%29)
subtract 2 from both sides
![y+2-2=\frac{1}{6}\left(x-6\right)-2](https://tex.z-dn.net/?f=y%2B2-2%3D%5Cfrac%7B1%7D%7B6%7D%5Cleft%28x-6%5Cright%29-2)
![y=\frac{1}{6}x-3](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B6%7Dx-3)
Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be: