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spayn [35]
2 years ago
5

What is the equation of the line that goes through the points (- 2, 1) and (0, 4) ? (Write your answer in standard form .)

Mathematics
1 answer:
tekilochka [14]2 years ago
5 0

Answer:

3x-2y+8=0

Step-by-step explanation:

The slope of the line that goes through the points (-2, 1) and (0, 4) is

\frac{4-1}{0-(-2)} = \frac{3}{2}

The slope of the line that goes through the points (x, y) and (0, 4) is

\frac{4-y}{0-x} = \frac{4-y}{-x}

So \frac{3}{2}=\frac{4-y}{-x}\\-3x=2(4-y)=8-2y\\3x-2y+8=0

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ExtremeBDS [4]

Answer:

x=\frac{3}{4}+i\frac{\sqrt{7}}{4},\:x=\frac{3}{4}-i\frac{\sqrt{7}}{4}

Step-by-step explanation:

simplify \frac{-3}{x-2} by putting the negative sign on the outside. \frac{2x}{x-1}-\frac{2x-5}{x^2-3x+2}=-\frac{3}{x-2}

find the LCM of the denominators. It is (x-1)(x-2). Multiply by the LCM:

\frac{2x}{x-1}\left(x-1\right)\left(x-2\right)-\frac{2x-5}{x^2-3x+2}\left(x-1\right)\left(x-2\right)=-\frac{3}{x-2}\left(x-1\right)\left(x-2\right)

Simplify:

2x\left(x-2\right)-\left(2x-5\right)=-3\left(x-1\right)

solve: x=\frac{3}{4}+i\frac{\sqrt{7}}{4},\:x=\frac{3}{4}-i\frac{\sqrt{7}}{4}

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