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how can this be written in parametric form?
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any thoughts? we have a point and a direction with any line
sorry I'm confused in this one
first step is just solving for y, put it into slope intercept form.
ok
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setiing: y = mx + b , then let x= 0 + 1t therefore the parametric version is x = t y = mt + b
the direction vector is (1,m) which can be reworked if we dont like fractions and the anchor point is (0,b)
i got y=5+(3x/2)
good, and let x=t and we have parametricised y in terms of t
thanks i got it :)
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i knew you would ;) good luck
x=t and y=5+(3t/2)
or equivalently x = 0 + 2t y = 5 + 3t if you dont like fractions
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