Find the slope of any line perpendicular to the line passing through (−7,3) and (−14,14)

omgespit9q

omgespit9q

Answered question

2022-10-25

Find the slope of any line perpendicular to the line passing through (−7,3) and (−14,14)

Answer & Explanation

cdtortosadn

cdtortosadn

Beginner2022-10-26Added 19 answers

The slope of any line perpendicular to another is the inverse of the slope of the reference line. The general line equation is y = mx + b, so the set of lines perpendicular to this will be y = -(1/m)x + c.

y = mx + b Calculate the slope, m, from the given point values, solve for b by using one of the point values, and check your solution using the other point values.

A line can be thought of as the ratio of the change between horizontal (x) and vertical (y) positions. Thus, for any two points defined by Cartesian (planar) coordinates such as those given in this problem, you simply set up the two changes (differences) and then make the ratio to obtain the slope, m.

Vertical difference “y” = y2 – y1 = 14 – 3 = 11
Horizontal difference “x” = x2 – x1 = -14 – -7 = -7
Ratio = “rise over run”, or vertical over horizontal = 11/-7 = -11/7 for the slope, m.

A line has the general form of y = mx + b, or vertical position is the product of the slope and horizontal position, x, plus the point where the line crosses (intercepts) the x-axis (the line where z is always zero.) So, once you have calculated the slope you can put any of the two points known into the equation, leaving us with only the intercept 'b' unknown.

3 = (-11/7)(-7) + b ; 3 = 11 + b ; -8 = b
Thus the final equation is y = -(11/7)x - 8

We then check this by substituting the other known point into the equation:

14 = (-11/7)(-14) - 8 ; 14 = 22 - 8 ; 14 = 14 CORRECT!

SO, if our original equation is y = -(11/7)x – 8, the set of lines perpendicular to it will have a slope of 7/11.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?