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mth 158
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** Question Form_labelMessages **
General form with one point
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If given point (1, -1) and slope -2/3
how do you find the general form?
the answer should be 2x + 3y = -1
But I don't understand how. Is there a formula?
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The slope from (1, -1) to (x, y) is
(y - 1) / (x - (-1))
which is the same as
(y - (-1)) / (x - 1).
If (x, y) lies on the line, then this slope must be -2/3, so an equation of the line woud be
(y -+1) / (x - 1) = -2/3
Solving for y, we multiply both sides by (x+1) to get
y + 1 = -2/3 ( x - 1)
then add 1 to both sides to get
y = -2/3 (x - 1) - 1.
Simplifying the right-hand side we get
y = -2/3 x + 2/3 - 1
or
y = -2/3 x - 1/3.
The general form is obtained by putting the x and y on the same side and having integer values for the coefficients of x and y. We do this by first multiplying through the equation by 3, obtaining
3 y = -2 x - 1
then adding 2x to both sides to get
2x + 3y = -1.
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To check our solution, we need to verify that (1, -1) is on the line, and that the line has slope -2/3.
The equation
y = -2/3 x - 1/3
is in slope-intercept form so the coefficient of x is the slope. That coefficient is -2/3, so the line does have slope -2/3.
If we substitute 1 for x and -1 for y we get
-1 = -2/3 * 1 - 1/3 or
-1 = -2/3 - 1/3 or
-1 = -1.
This verifies that (1, -1) lines on the line.
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