The line parallel to $3 x + 4 y + 7 = 0$ and passing through $(1, 2)$ is ____

3 x + 4 y + 11 = 0

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4 x - 3 y + 11 = 0

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3 x + 4 y - 11 = 0

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x + y - 3 = 0

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Solution

The correct option is C $3 x + 4 y - 11 = 0$
Two parallel lines have the same slope, put $3 x + 4 y = - 7$ in slope-intercept form:
$3 x + 4 y = - 7$
$4 y = - 3 x - 7$
$y = - \frac{3}{4} x - \frac{7}{4}$
The slope is $- \frac{3}{4}$ , so a line parallel to this has a slope of $- \frac{3}{4}$ also.
Now since the equation of a line is $y - y_{1} = m (x - x_{1})$ , where $m$ is the slope of the line, therefore with point $(1, 2)$
We have, $(y - 2) = - \frac{3}{4} (x - 1) \Rightarrow 4 (y - 2) = - 3 (x - 1) \Rightarrow 4 y - 8 = - 3 x + 3 \Rightarrow 3 x + 4 y - 8 - 3 = 0 \Rightarrow 3 x + 4 y - 11 = 0$
Hence, resulting linear equation is $3 x + 4 y - 11 = 0$ .

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