Question

Find the equation of the line that is perpendicular to y=-(2)/(7)x+9 and passes through the point (4,−6).

Answer 1

The equation of the line that is perpendicular to y=-(2/7)x+9 and passes through the point (4,-6) is y = (7/2)x - 20.

Step-by-step explanation:

We determine the negative reciprocal of the slope of the given line, which is 7/2. Using the point-slope form of a line, we substitute the values into the equation y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope. Simplifying the equation results in y = (7/2)x - 20, which is the equation of the perpendicular line passing through (4,-6).

Answer 2

The equation of the line perpendicular to y=-(2/7)x+9 and passes through the point (4, -6) is y = (7/2)x - 20.

Step-by-step explanation:

To find the equation of a line that is perpendicular to another line and passes through a given point, we first need to determine the slope of the original line. The given line is y=-(2/7)x+9, so its slope is -(2/7). Lines that are perpendicular to one another have slopes that are negative reciprocals of each other. So, the slope of the line we are looking for will be 7/2.

Now, using the point-slope form of a line equation, which is y - y1 = m(x - x1), and substituting the slope we found (7/2) and the point given (4, -6), we get:

y - (-6) = (7/2)(x - 4)

y + 6 = (7/2)x - 14

Finally, subtract 6 from both sides to solve for y:

y = (7/2)x - 14 - 6

y = (7/2)x - 20

So, the equation of the line perpendicular to y=-(2/7)x+9 and passing through the point (4, -6) is y = (7/2)x - 20.