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  • one year ago

Help!! Write the equation of a line, in slope-intercept form, that is parallel to the line -12x+3y=23 , and passes through the point (2,-3).

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  1. texaschic101
    • one year ago
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    first we need to find the slope in the original equation because parallel lines have the same slope. We will put the equation in y = mx + b form where m is your slope. -12x + 3y = 23 --- add 12x to both sides 3y = 12x + 23 -- now divide both sides by 3 y = 4x + 23/3 now the number in the m position, the slope, is 4 Our parallel line will have a slope of 4. Now we will use y = mx + b again... slope(m) = 4 (2,-3)...x = 2 and y = -3 and since we have our slope and our points (x,y), we just need to find b, the y intercept. now we sub -3 = 4(2) + b -3 = 8 + b -8-3 = b -11 = b so your parallel equation is : y = 4x + (-11) or y = 4x - 11 questions ?

  2. anonymous
    • one year ago
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    @texaschic101 Thanks and how about this ? Write the equation, in slope-intercept form, that is perpendicular to y=-1/2x-7 that passes through the point (-4, 5).

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