Find the distance between parallel lines
(i) \(15x + 8y – 34 = 0\) and \(15x + 8y + 31 = 0\)
(ii) \(l (x + y) + p = 0\) and \(l (x + y) – r = 0.\)
(i) \(15x + 8y – 34 = 0\) and \(15x + 8y + 31 = 0\)
(ii) \(l (x + y) + p = 0\) and \(l (x + y) – r = 0.\)
Solution and Explanation
It is known that the distance (d) between parallel lines \(Ax + By + C1 = 0\) and \(Ax + By + C2 = 0 \) is given by
\(d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}\)
(i) The given parallel lines are \(15x + 8y - 34 = 0\) and \(15x + 8y + 31 = 0.\)
Here, \(A = 15, B = 8, C1 = -34\), and \(C2 = 31\).
Therefore, the distance between the parallel lines is
\(d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}\)
\(=\frac{|-34-31|}{\sqrt{(15)^2+(8)^2}}\) units
\(=\frac{|-65|}{17}\) units
\(=\frac{65}{17} \) units.
(ii) The given parallel lines are \( l (x + y) + p = 0\) and \(l (x + y) - r = 0.\)
\(lx + ly + p = 0\) and \( lx + ly - r = 0 \)
Here, \(A = l, B = l, C_1 = p\), and \(C_2 = - r.\)
Therefore, the distance between the parallel lines is
\(d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}\)
\(=\frac{|p+r|}{\sqrt{l^2+l^2}}\) units
\(=\frac{|p+r|}{\sqrt{2l^2}}\) units
\(=\frac{|p+r|}{l\sqrt{2}} \) units
\(=\frac{1}{\sqrt{2}}|\frac{p+r}{l}|\) units.
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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