$\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} $

- (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,

$\qquad 4y-12=3x+6$

$\quad \Rightarrow 3x-4y+18=0$

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