# 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.

Algebra ->  Equations -> 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.      Log On

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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(60776)   (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. -------------- The given line has slope = (-1/3) --- Any line perpendicular to it must have slope = 3 ---- Find the y-intercept: 2 = 3(-9)+b 2 = -27 + b b = 29 --------------- Equation: y = 3x + 29 ============================ Cheers, Stan H. Answer by MathTherapy(1817)   (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