Maths question year 10
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Maths question year 10

[From: ] [author: ] [Date: 12-03-20] [Hit: ]
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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

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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

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Use point slope form!
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
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