Find the equation of a straight line that passes through point (6,2) and has a gradient of 4.
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Using the equation : (y - y') = m(x - x')
Here y' = 2 ; x' = 6 ; m = 4
(y - 2) = 4(x - 6)
y - 2 = 4x - 24
4x -y -22 = 0 --------------> Required equation
Here y' = 2 ; x' = 6 ; m = 4
(y - 2) = 4(x - 6)
y - 2 = 4x - 24
4x -y -22 = 0 --------------> Required equation
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We can find the equation by using
y - y1 = m (x - x1)
here m = 4 and (x1, y1) = (6,2)
so y - 2 = 4 ( x - 6)
y - 2 = 4x -24
y = 4x -24 +2
y = 4x -22
or 4x - y - 22 = 0
y - y1 = m (x - x1)
here m = 4 and (x1, y1) = (6,2)
so y - 2 = 4 ( x - 6)
y - 2 = 4x -24
y = 4x -24 +2
y = 4x -22
or 4x - y - 22 = 0
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Use point slope form!
y-2= 4(x-6)
Simplify!
y-2= 4(x-6)
Simplify!
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y=mx+c
sub in points and gradient 2=(4)(6)+c
c=-22
y=4x-22
sub in points and gradient 2=(4)(6)+c
c=-22
y=4x-22