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Equation of line passing through the point (1, 2) and perpendicular to the line y = 3x - 1 is A. x+3y+7=0 B. x+3y-7=0 B.x+3y=0 D. x-3y=0

User BenRoob
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1 Answer

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To find the equation of the line that passes through the point (1, 2) and is perpendicular to the line y=3x-1, we can first find the slope of the line y=3x-1. The slope of this line is 3, so a line perpendicular to it will have a slope of -1/3. We can then use the point-slope form of the line to write the equation of the line that passes through (1,2) and has a slope of -1/3. The point-slap form of a line is given by y-y₁ = m(x-x₁), where (x₁,y₁) is a point on the line and m is the slope of the line. Plugging in the coordinates of the point (1,2) and the slope -1/3, we get the equation y-2 = -1/3(x-1), which simplifies to y-2 = -x/3 + 1/3. Multiplying both sides of the equation by 3 to get rid of the fractions, we get 3y-6 = -x+1. Finally, we can move all the terms to one side of the equation to get the standard form of the line: 3y-x-5 = 0. Therefore, the equation of the line that passes through the point (1, 2) and is perpendicular to the line y = 3x - 1 is x-3y-5=0, so the correct answer is A.

User LeoMestizo
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