Find Equation of the Line
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Find Equation of the Line

[From: ] [author: ] [Date: 12-08-02] [Hit: ]
so plug x and the y-intercept into the form y=ax+b.y=x/(3sqrt(2))+sqrt(2) or y=x/(18^(1/2))+2^(1/2), depending on which square root notation you prefer.-Are these two separate equations you want?......
Find an equation for the line that is described and make sure your final answer is in the form y = mx + b.

(1) Passes through (6,2) and has the same x-intercept as the line -2x + y = 1

(2) Has x-intercept -6 and y-intercept square root 2

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(1) the x intercept of the first answer is where y=0, or where -2x+0=1, so it's at x=-1/2. (-1/2,0) to (6,2) has a rise of 2 and a run of 6.5. Slope is equal to rise/run, or 2/6.5, which can be simplified to 4/13.
Slope=4/13, so m=4/13. multiply this by your given x-coordinate and subtract that from the given 7-coordinate.
2-(4/13)*6, or 2/13.
So the line in the form of y=ax+b is y=4x/13+2/13.
NOTE: "square root 2" is generally written "sqrt(2)" or "2^(1/2)," whichever is appropriate.
(2) x-intercept=-6, y-intercept=sqrt(2). Find the slope first with sqrt(2)/-6=1/(3sqrt(2)). You were already given the y-intercept, so plug x and the y-intercept into the form y=ax+b.
y=x/(3sqrt(2))+sqrt(2) or y=x/(18^(1/2))+2^(1/2), depending on which square root notation you prefer.

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Are these two separate equations you want?
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