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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} $

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  • (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,
$\qquad 4y-12=3x+6$
$\quad \Rightarrow 3x-4y+18=0$
answered May 10, 2014 by thanvigandhi_1

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