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

$\begin{array}{1 1}(A)\;3x-4y-18=0 \\(B)\; 3x-4y+18=0 \\(C)\; 4x-3y-18=0 \\(D)\;4x-3y+18=0 \end{array}$

Toolbox:
• (a) Equation of a line in the slope intercept form is $y=mx+c$
• (b) Parallel lines have same slope.
• (c) Equation of a line with given slope and a point $(x_1, y_1)$ is $y-y_1=m(x-x_1)$
The equation of the line given is
$\qquad 3x-4y+2=0$
This can be written as
$\qquad 4y=3x+2$
$\quad \Rightarrow y=\large\frac{3}{4}$$x+\large\frac{1}{2} This is of the form y=mx+c It is clear that m=\large\frac{3}{4} and c=\large\frac{1}{2} Since the lines are parallel then slopes are equal. Hence the slope of the other line is also \large\frac{3}{4} Hence equation of the line with slope \large\frac{3}{4} and passing through (-2,3) is \qquad y-3=\large\frac{3}{4}$$(x-(-2))$
$\quad \Rightarrow 4(y-3)=3(x+2)$
On simplifying we get,
$\qquad 4y-12=3x+6$
$\quad \Rightarrow 3x-4y+18=0$