(A) 2x + 5y = 7
(B) 2x − 5y = 7
(C) 5x + 2y = 7
(D) 5x − 2y =10
(E) 5x − 5y =10
Don't worry about me cheating. I already know the answer is A due to me having the answer sheet. However, I want to know how to work out this problem so that when I am faced with something similar I will know what to do.
How do I solve this: In the xy-plane, the line given by which of the following is perpendicular to the line 5x − 2y = 7 ?
2020-04-05 1:18 pm
回答 (5)
2020-04-05 6:00 pm
✔ 最佳答案
5x - 2y = 7 can be re-written as:2y = 5x - 7
i.e. y = 5x/2 - 7/2
Hence, gradient 5/2
Now, a line perpendicular to this line will have gradient -2/5
so, y = -2x/5 + C....where C is a constant
or, 5y = -2x + 5C
i.e. 5y + 2x = 5C
Then, we have 5y + 2x = 7....if C = 7/5
:)>
2020-04-05 1:38 pm
Given line: 5x - 2y = 7
Slope of the given line = -5/(-2) = 5/2
Slope of a line perpendicular to the given line = -2/5
Equation of the line: y = (-2/5)x + c, i.e.
5y = -2x + 5c
2x + 5y = 5c
The answer: (A) 2x + 5y = 7
Slope of the given line = -5/(-2) = 5/2
Slope of a line perpendicular to the given line = -2/5
Equation of the line: y = (-2/5)x + c, i.e.
5y = -2x + 5c
2x + 5y = 5c
The answer: (A) 2x + 5y = 7
2020-04-06 12:40 am
Put it into slope intercept form Y=Mx+B
-2y=-5x+7, 2y=5x-7, y=5x/2 -7/2.
The neg inverse of the x term is Perpendicular.
Look here: https://www.khanacademy.org/math/geometry/hs-geo-analytic-geometry/hs-geo-parallel-perpendicular-eq/v/perpendicular-lines-2
-2y=-5x+7, 2y=5x-7, y=5x/2 -7/2.
The neg inverse of the x term is Perpendicular.
Look here: https://www.khanacademy.org/math/geometry/hs-geo-analytic-geometry/hs-geo-parallel-perpendicular-eq/v/perpendicular-lines-2
2020-04-05 9:21 pm
The line perpendicular to ax+by=c through a given point (x₀,y₀) is bx - ay = bx₀ - ay₀.
Here this will be 2x + 5y = ...
Here this will be 2x + 5y = ...
2020-04-05 1:31 pm
There are lots of Youtube videos if you are actually interested in learning, I suspect not.
收錄日期: 2021-04-18 18:31:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200405051800AAy36kB