Find equation of the line which is equidistant from parallel lines 9x + 6y - 7 = 0 and 3x + 2y + 6 = 0.
Solution and Explanation
The equations of the given lines are
\(9x + 6y - 7 = 0 … (1) \)
\(3x + 2y + 6 = 0 … (2) \)
Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2).
The perpendicular distance of P (h, k) from line (1) is given by
\(d_1=\frac{\left|9h+6k-7\right|}{\sqrt{(9)^2+(6)^2}}\)
\(=\frac{\left|9h+6k-7\right|}{\sqrt{117}}\)
\(=\frac{\left|9h+6k-7\right|}{3\sqrt{13}}\)
The perpendicular distance of P (h, k) from line (2) is given by
\(d_2=\frac{\left|3h+2k+6\right|}{\sqrt{(3)^2+(2)^2}}\)
\(=\frac{\left|3h+2k+6\right|}{\sqrt{13}}\)
Since P (h, k) is equidistant from lines (1) and (2), \(d_1=d_2\)
\(∴\) \(\frac{\left|9h+6k-7\right|}{3\sqrt{13}}\)\(=\frac{\left|3h+2k+6\right|}{\sqrt{13}}\)
\(⇒ |9h+6k-7|=3|3h+2k+6|\)
\(⇒ |9h+6k-7|=±3(3h+2k+6)\)
\(⇒ 9h+6k-7=3(3h+2k+6)\) or \(9h+6k-7=-3(3h+2k+6) \)
The case \(9h+6k-7=3(3h+2k+6)\) is not possible.
\(9h+6k-7=3(3h+2k+6) ⇒ -7=18\) (which is absurd)
\(∴ 9h+6k-7=-3(3h+2k+6) \)
\(9h + 6k - 7 = - 9h - 6k - 18 \)
\(⇒ 18h + 12k + 11 = 0\)
Thus, the required equation of the line is \(18x + 12y + 11 = 0.\)
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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