Find the equation of the line parallel to the | vedclass

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(A) The equation of the given line is $3x - 4y + 2 = 0$.The slope-intercept form of a line is $y = mx + c$.Rewriting the given equation: $4y = 3x + 2$

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$3x - 4y + 18 = 0$

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(A) The equation of the given line is $3x - 4y + 2 = 0$.The slope-intercept form of a line is $y = mx + c$.Rewriting the given equation: $4y = 3x + 2$,which gives $y = \frac{3}{4}x + \frac{1}{2}$.Thus,the slope $(m)$ of the given line is $\frac{3}{4}$.Since parallel lines have the same slope,the slope of the required line is also $m = \frac{3}{4}$.Using the point-slope form $(y - y_1) = m(x - x_1)$ for the point $(-2, 3)$:$(y - 3) = \frac{3}{4}(x - (-2))$$4(y - 3) = 3(x + 2)$$4y - 12 = 3x + 6$$3x - 4y + 18 = 0$.

Find the equation of the line parallel to the line $3x - 4y + 2 = 0$ and passing through the point $(-2, 3).$

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