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What is the equation in standard form of the line which passes through (4, –2) and has a slope of –3? OPTIONS: A)x + 3y = 10 B) x + 3y = –10 C) 3x + y = 10 D) 3x + y = –10
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\(\bf \begin{array}{lllll} &x_1&y_1\\ &(4\quad ,&-2) \end{array} \\\quad \\ slope = m= -3 \\ \quad \\ y-y_1=m(x-x_1)\quad \textit{plug in your values and solve for "y"}\)
i dont get how to solve foy y
well, like you did on the previous one, you isolate "y"
hmmm actually the choices show that you don' t have to :/
y-(-2)=-3(x-4) y+2=-3x+12 y=3x+10
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y -(-2) = -3 ( x - 4) y+2 = -3x+12 3x + y = 10
okay thanks
pretty much, yes, yours is correct, just missed the - in the last line
y-(-2)=-3(x-4) y+2=-3x+12 y=3x+10
i dont get how to solve foy y
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