Equation of the line passing through (1, 2) and parallel to the line y = 3x – 1 is
A
y + 2 = x + 1
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B
y + 2 = 3(x + 1)
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C
y – 2 = 3(x – 1)
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D
y – 2 = x – 1
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Solution
The correct option is C y – 2 = 3(x – 1) In the equation y = 3x – 1 the slope of the line paralle to this line will be 3. Therefore the required equation which passes through (1, 2) and whose gradient is 3, is (y – 2) = 3(x – 1).