Find the equation of a straight line which cuts off an intercept of length 3 on y - axis and is parallel to the line joining the points (3, -2) and (1, 4).

3x + y - 3 = 0

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3x + y - 1 = 0

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3x + 2y - 3 = 0

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3x + y - 7 = 0

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Solution

The correct option is A 3x + y - 3 = 0
Given; c = 3, Slope (m) $= \frac{4 + 2}{1 - 3} = - 3$
So, equation of line is $y = - 3 x + 3 \Rightarrow 3 x + y - 3 = 0$

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Find the equation of a straight line which cuts off an intercept of length 3 on y - axis and is parallel to the line joining the points (3, -2) and (1, 4).

The correct option is A 3x + y - 3 = 0Given; c = 3, Slope (m) =4+21−3=−3 So, equation of line is y=−3x+3⇒3x+y−3=0

The correct option is A 3x + y - 3 = 0
Given; c = 3, Slope (m) $= \frac{4 + 2}{1 - 3} = - 3$
So, equation of line is $y = - 3 x + 3 \Rightarrow 3 x + y - 3 = 0$