SOLUTION: Please help me find the solution to this problem:
Find the equation of a straight line through (-3,2) that is perpendicular to X+3y=6 with the equation in standard form.
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Question 204911: Please help me find the solution to this problem:
Find the equation of a straight line through (-3,2) that is perpendicular to X+3y=6 with the equation in standard form.
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the equation of a straight line through (-3,2) that is perpendicular to X+3y=6 with the equation in standard form.
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The given line has slope = (-1/3)
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Any line perpendicular to it must have slope = 3
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Find the y-intercept:
2 = 3(-9)+b
2 = -27 + b
b = 29
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Equation: y = 3x + 29
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Cheers,
Stan H.
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
x + 3y = 6
3y = -x + 6
Since the equation of the line that passes through the point (-3, 2) is perpendicular to: , then its slope is 3
Therefore, using the point-slope formula: , we get:
y - 2 = 3(x - -3)
y - 2 = 3x + 9
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