How do you find a standard form equation for the line (0,2) (1,-2)?

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How do you find a standard form equation for the line (0,2) (1,-2)?

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Answer 1 · 2018-01-09T02:53:31

4x + y = 2

Explanation:

Your first step in this process is to find the slope.
$\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $(y_"2"-y_"1")/(x_"2"-x_"1")$ = $\frac{- 2 - 2}{1 - 0}$ $(-2-2)/(1-0)$

This gives you a slope of $- \frac{4}{1}$ $-4/1$

Your second step is to plug it into the point slope form with one of the points you were given.
$(y - y_{1}) = m (x - x_{1})$ $(y-y_"1")=m(x-x_"1")$ = $(y - 2) = - \frac{4}{1} (x - 0)$ $(y-2)=-4/1(x-0)$

Your third step is to simplify that into your standard form which is $A x + B x = C$ $Ax+Bx=C$
$1 (y - 2) = 1 [- 4 (x - 0)]$ $1(y-2) = 1[ -4(x-0)]$
$y - 2 = - 4 x + 0$ $y-2=-4x+0$ Distribute
$y - 2 + 4 x = 0$ $y-2+4x=0$ Move the $x$ $x$ value to the other side with inverse operations.
$y + 4 x = 2$ $y+4x=2$ Move the $C$ $C$ value to the other side with inverse operations.
$4 x + y = 2$ $4x+y=2$ Rearrange accordingly to get into the $A x + B y = C$ $Ax + By=C$ formula.

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How do you find a standard form equation for the line (0,2) (1,-2)?

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