Find the equation of the line that passes through (1,3) and is perpendicular to y=1-2x Leave your answer in the form y=mx=c?
回答 (8)
2019-10-12 7:48 pm
y = (-1/(-2))x + 3 + 1/(-2) which simplifies to y = ½x + ⁵∕₂
參考: The line perpendicular to y=mx+b through a given point (x₀,y₀) is y = (-1/m)x + y₀ + x₀/m.
2019-10-14 7:23 pm
y = 1 – 2x ... (1)
perpendicular line is x-2y+k = 0it passing through (1, 3) → 1-2 x 3 + k = 0required equation is x-2y+5 = 0.
perpendicular line is x-2y+k = 0it passing through (1, 3) → 1-2 x 3 + k = 0required equation is x-2y+5 = 0.
2019-10-13 6:50 pm
Slope of perpendicular is 1/2.
y = (1/2)x + b
3 = (1/2)1 + b
5/2 = b
y = (1/2)x + 5/2
y = (1/2)x + b
3 = (1/2)1 + b
5/2 = b
y = (1/2)x + 5/2
2019-10-12 7:26 pm
The most direct way to solve this is to find the slope of desired line, then use the point-slope formula for the line with that slope through the given point.
The given line has slope -2, so any line perpendicular to that will have slope m=(-1)/(-2) = 1/2.
y - b = m(x - a) . . . . point-slope equation for line through (a,b) with slope m
y - 3 = (1/2)(x - 1) . . . . substitute (a,b) = (1,3) and m = 1/2
Then express that in whatever form you need. For slope-intercept form, just add 3 and simplify
y = (1/2) x - 1/2 + 3
y = (1/2) x + 5/2
The given line has slope -2, so any line perpendicular to that will have slope m=(-1)/(-2) = 1/2.
y - b = m(x - a) . . . . point-slope equation for line through (a,b) with slope m
y - 3 = (1/2)(x - 1) . . . . substitute (a,b) = (1,3) and m = 1/2
Then express that in whatever form you need. For slope-intercept form, just add 3 and simplify
y = (1/2) x - 1/2 + 3
y = (1/2) x + 5/2
2019-10-12 7:46 pm
Perpendicular slope: 1/2
Perpendicular equation: y = 1/2x + 5/2
Perpendicular equation: y = 1/2x + 5/2
2019-10-12 7:21 pm
Find the equation of the line that passes through (1,3) and is perpendicular to y=1-2x Leave your answer in the form y=mx=c?
assuming that is y = mx+c
y = -2x + 1
slope is –2 and perpendicular would be a slope of +1/2
so equation is
y = (1/2)x + k
use point to determine k
3 = (1/2)1 + k
k = 3 – (1/2) = 5/2
so equation is
y = (1/2)x + (5/2)
assuming that is y = mx+c
y = -2x + 1
slope is –2 and perpendicular would be a slope of +1/2
so equation is
y = (1/2)x + k
use point to determine k
3 = (1/2)1 + k
k = 3 – (1/2) = 5/2
so equation is
y = (1/2)x + (5/2)
2019-10-12 7:21 pm
For the line (y = 1 - 2x), the slope = -2
The required line is perpendicular to (y = 1 - 2x).
The slope of the required line = -1/(-2) = 1/2
The equation of the required line (point-slope form):
(y - 3) = (1/2)(x - 1)
2(y - 3) = x - 1
2y - 6 = x - 1
2y = x + 5
y = (1/2)x + (5/2)
The required line is perpendicular to (y = 1 - 2x).
The slope of the required line = -1/(-2) = 1/2
The equation of the required line (point-slope form):
(y - 3) = (1/2)(x - 1)
2(y - 3) = x - 1
2y - 6 = x - 1
2y = x + 5
y = (1/2)x + (5/2)
2019-10-12 7:19 pm
y = 1 - 2x has a slope of -2.
The perpendicular line will have a slope of -1/(-2) = 1/2.
The perpendicular line is
y - 3 = 1/2(x - 1)
y - 3 = x/2 - 1/2
y = x/2 + 5/2
The perpendicular line will have a slope of -1/(-2) = 1/2.
The perpendicular line is
y - 3 = 1/2(x - 1)
y - 3 = x/2 - 1/2
y = x/2 + 5/2
收錄日期: 2021-04-18 18:21:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20191012111015AApyGr2