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

Question

If a point $ P(x, y) $ satisfies the condition that its distance from the point $ (3, -2) $ is equal to its distance from the line $ y = 2x + 1 $, then the locus of point $ P $ is:

Answer 1

The equation of the given line is
$3x – 4y + 2 = 0$

or $y = \frac{3x}{4} +\frac{ 2}{4}$

$y=\frac{ 3x}{4} + \frac{1}{2}$, which is of the form $y = mx + c $

∴ Slope of the given line=$\frac{3}{4}$
It is known that parallel lines have the same slope.

∴ Slope of the other line $=m=\frac{3}{4}$
Now, the equation of the line that has a slope of $\frac{3}{4}$ and passes through the point (-2, 3) is
$y – 3 = \frac{3}{4} (x – (-2))$
$4y – 12 = 3x + 6$
$i.e,3x – 4y + 18 = 0$

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

Solution and Explanation

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