What is the equation that contains point -4,1 and is parallel to x+2y=3?
回答 (2)
2011-04-08 5:58 pm
slope of the required equation
= slope of x + 2y = 3 [ Ax + Bx + C = 0 ]
= -1/2 [ slope = - A / B ]
required equation is
(y - 1) / (x - (-4)) = - 1/2
(y - 1) / (x + 4) = - 1/2
- (x + 4) = 2 (y - 1)
- x - 4 = 2y - 2
- x -4 -2y + 2 = 0
- x -2y -2 = 0
x + 2y + 2 = 0
= slope of x + 2y = 3 [ Ax + Bx + C = 0 ]
= -1/2 [ slope = - A / B ]
required equation is
(y - 1) / (x - (-4)) = - 1/2
(y - 1) / (x + 4) = - 1/2
- (x + 4) = 2 (y - 1)
- x - 4 = 2y - 2
- x -4 -2y + 2 = 0
- x -2y -2 = 0
x + 2y + 2 = 0
2011-04-08 4:35 pm
2y = - x + 3
y = (-1/2) x + 3/2
m = - 1/2
y - 1 = (-1/2) ( x + 4 )
y = (-1/2) x - 1
y = (-1/2) x + 3/2
m = - 1/2
y - 1 = (-1/2) ( x + 4 )
y = (-1/2) x - 1
收錄日期: 2021-05-01 00:08:44
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