How do you find the slope of a line perpendicular to #y=7/5x-2#?

Question

How do you find the slope of a line perpendicular to #y=7/5x-2#?

↳Redirected from "How do I change #int_0^1int_0^sqrt(1-x^2)int_sqrt(x^2+y^2)^sqrt(2-x^2-y^2)xydzdydx# to cylindrical or spherical coordinates?"

1 Answer

Impact of this question

2451 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-20T11:32:38

The slope is #-5/7#

Explanation:

The slope of your line is #7/5#, therefore a director vector of your line is, for example, #vec{AB} (5,7)# (because the slope is #(Delta y) /(Delta x)#).

The vector #vec{u} (-7,5)# is orthogonal with #vec{AB}# because dot product is null : #5 \times(-7) + 7\times 5 = 0#.

Finally, the slope is #(Delta y) /(Delta x) = 5/-7 = - 5/7#.

Remark. You need orthonormal system for the equation of the line. But I suppose it is :-)

Science

Math

Humanities

... and beyond

How do you find the slope of a line perpendicular to #y=7/5x-2#?

1 Answer

Explanation:

Related questions

Impact of this question