Find equation of the line parallel to the line \(3x - 4y + 2 = 0\) and passing through the point (-2, 3).
Solution and Explanation
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
\(y – 3 = \frac{3}{4} (x – (-2))\)
\(4y – 12 = 3x + 6\)
\(i.e,3x – 4y + 18 = 0\)
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Concepts Used:
Straight lines
A straight line is a line having the shortest distance between two points.
A straight line can be represented as an equation in various forms, as show in the image below:
The following are the many forms of the equation of the line that are presented in straight line-
1. Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.
y – y0 = m (x – x0)
2. Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
The slope of P2P = The slope of P1P2 , i.e.
\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)
Hence, the equation becomes:
y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
3. Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by
y – c =m( x - 0 )
As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c