Describe how to graph a line given an equation in standard form.
1 answer:
This is pretty simple. So a standard form equation is probably something like this: 2x+8y=10.
We have to get that equation right there into a slope intercept, which is just isolating the variable y.
2x+8y=16
-2x -2x Subtract the x value.
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(8y=16-2x)/8 Divide the whole equation by 8 to isolate the y value.
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y=2-1/4x
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The fraction with the variable x is the slope. In this case, the slope is -1/4, so meaning it goes down one unit and goes right 4.
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The 2 is the y-intercept. (0, 2)
We have to get that equation right there into a slope intercept, which is just isolating the variable y.
2x+8y=16
-2x -2x Subtract the x value.
------------------
(8y=16-2x)/8 Divide the whole equation by 8 to isolate the y value.
------------------
y=2-1/4x
------------------
The fraction with the variable x is the slope. In this case, the slope is -1/4, so meaning it goes down one unit and goes right 4.
------------------
The 2 is the y-intercept. (0, 2)
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Given:
A line through the points (7,1,-5) and (3,4,-2).
To find:
The parametric equations of the line.
Solution:
Direction vector for the points (7,1,-5) and (3,4,-2) is
Now, the perimetric equations for initial point with direction vector
, are
The initial point is (7,1,-5) and direction vector is . So the perimetric equations are
Similarly,
Therefore, the required perimetric equations are and
.