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anonymous

  • 5 years ago

How do you write the equation of the line parallel to y=3x-1 that passes through the point (-3, -5)?

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  1. anonymous
    • 5 years ago
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    parallel means identical slopes, so m=3, and using pt-slope form, Y-Y1=m(x-x1), the parallel line is Y+5=3(X+3) so Y=3X+9-5 So........................Y=3X+4 Y=3X+4 is the eqtn.

  2. anonymous
    • 5 years ago
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    Wow. Dude can um you explain that to me?

  3. shadowfiend
    • 5 years ago
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    So, two lines are only parallel if they have the same slopes. You can write a simple line in the form y = mx + b, where m is the slope and b is the y-intercept -- the y value the graph of the function runs through when x = 0.

  4. shadowfiend
    • 5 years ago
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    You are given one function, which is \(y = 3x - 1\). This tells you that the slope for the second line must be 3 in order for it to be parallel. The only thing you have left to find out then is the y-intercept that will let the line go through the point (-3, -5). A point is a pair, (x, y) that gives you an X value and Y value that are on the line. Since you have m, you can say: $$ y = 3x + b $$ $$ (-5) = 3 \cdot (-3) + b $$ Here we plugged in the point that we want to be on the line -- (-3, -5) -- into our equation. Then, we can solve for b: $$ -5 - (3 \cdot (-3)) = b $$ This will give us the value of b, which we can then plug back into the equation: $$ y = 3x + (-5 - (3 \cdot (-3))) $$

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