What would be. perpendicular to y equals -3x-2?

Answer 1

y = 1/3x + c

Alternate Answer:

Any line with a slope of m is perpendicular to any line with a slope of 1/-m.

Proof:

We are working in the Euclidean space of R2. A vector fis orthogonal to another vector g, if their dot product is 0. The dot product is: fTg, or the sum of the product of the corresponding elements in each vector. I.e. fTg = f1*g1 + f2*g2 + f3g3 + ... + fngn

First, Find the vector equation of a line centered at the origin:

y = mx

plug in x = 1

y = m

This give you the direction vector m = (1, m)

This in turn gives us the vector equation:

r = mx

r = (1, m)*x

We want to find some m', such that mTm'= 0

Let m' = (c,d). We want to find c and d, such that 1c + md = 0

c = -md

Therefore, m' = (-md,d), with d being any real number.

Thus, the equation

r' = m'*x

is orthogonal/perpendicular to r

Finally, we want to convert back to linear equations for both of our vector equations.

We can express a vector equation of the form,

s = (a, b)*x

as a linear equation:

a*y = b*x

if we substitute r' for s, and m' for (a, b), we have:

-m*d*y = d*x

by isolating the y variable, we have:

y = (1/-m)*x

which is orthogonal to the line:

y = m*x

Now, we are almost done.

For a line y = mx + b, we should note the angle it forms with another line, y = m'x + b' is always the same, for fixed m and m', and arbitrary b and b'. Thus the values of b and b' do not change the orthogonality of the system.

QED.

SO, for YOUR particular example, we would have y = -3x - 2 being orthogonal/perpendicular to ALL lines of the form y = (1/3)x + b, for any real number b.