Question

Given a line with a slope of m = 6 and passes y-intercept (O, 1), determine the equation of the through the point (-2, 6), determine the equation line perpendicular to the given line that passes through the point (6, -4). Represent the line in of the line perpendicular to the given line that slope-intercept and standard forms. passes through the point (-9, 1). Represent the line in slope-intercept and standard forms.

Answer 1

Using the point-slope form, the slope-intercept form of the equation is y = -1/6x - 3, which can also be represented in standard form as x + 6y = -18.

Step-by-step explanation:

To find the equation of the line perpendicular to a given line with a slope of m = 6, which passes through the point (6, -4), we first need to determine the slope of the perpendicular line. The slope of any line perpendicular to another line is the negative reciprocal of the original line's slope. Therefore, for our line, the slope will be m = -1/6.

Now, using the point-slope form of the linear equation, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line ((6, -4) in our case), we can plug in our values:

y - (-4) = -1/6(x - 6)

This simplifies to:

y + 4 = -1/6x + 1

Then, to get the slope-intercept form (y = mx + b), we subtract 4 from both sides:

y = -1/6x - 3

For the standard form (Ax + By = C), we multiply both sides by 6 to eliminate the fraction and move the terms involving x and y to one side:

6y = -x - 18

Add x to both sides to get:

x + 6y = -18

So, the standard form of the equation is x + 6y = -18.