Question

What is an equation of the line that passes through the point ( 6 , − 2 ) (6,−2) and is perpendicular to the line 6 x + y = 2 6x+y=2?

Answer 1

Answer:

An equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:

• $y=\frac{1}{6}x-3$

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

$y-y_1=m\left(x-x_1\right)$

substituting the values m = 1/6 and the point (6, -2)

$y-\left(-2\right)=\frac{1}{6}\left(x-6\right)$

$y+2=\frac{1}{6}\left(x-6\right)$

subtract 2 from both sides

$y+2-2=\frac{1}{6}\left(x-6\right)-2$

$y=\frac{1}{6}x-3$

Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:

• $y=\frac{1}{6}x-3$