the equation of the line that passes through (1,1) and (3,-4) is? ?
回答 (3)
2020-08-19 12:26 pm
The equation of the line (two-point form):
(y - 1)/(x - 1) = (-4 - 1)/(3 - 1)
(y - 1)/(x - 1) = -5/2
2(y - 1) = -5(x - 1)
2y - 2 = -5x + 5
5x + 2y - 7 = 0
(y - 1)/(x - 1) = (-4 - 1)/(3 - 1)
(y - 1)/(x - 1) = -5/2
2(y - 1) = -5(x - 1)
2y - 2 = -5x + 5
5x + 2y - 7 = 0
2020-08-19 9:46 pm
Points: (1, 1) and (3, -4)
Slope: -5/2 or -2.5
Equation: y = -2.5x +3.5
Slope: -5/2 or -2.5
Equation: y = -2.5x +3.5
2020-08-19 7:25 pm
Gradient is (-4 - 1)/(3 - 1) = -5/2
so, y = -5x/2 + C
Using point (1, 1) we have:
1 = -5/2 + C
Hence, C = 7/2
so, y = -5x/2 + 7/2
i.e. 2y = -5x + 7
or, 2y + 5x - 7 = 0...in standard form
:)>
so, y = -5x/2 + C
Using point (1, 1) we have:
1 = -5/2 + C
Hence, C = 7/2
so, y = -5x/2 + 7/2
i.e. 2y = -5x + 7
or, 2y + 5x - 7 = 0...in standard form
:)>
收錄日期: 2021-04-24 07:58:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200819042317AA6bVZY