Find angles between the lines \(\sqrt3x+y=1\) and \(x+\sqrt3y=1.\)
Solution and Explanation
The given lines are \(\sqrt3x+y=1\) and \(x+\sqrt3y=1.\)
\(y = -\sqrt3x + 1 … (1)\) and \(y = \frac{-1}{\sqrt3x} +\frac{ 1}{\sqrt3} …. (2)\)
The slope of line (1) is \(m_1 = -\sqrt3\) , while the slope of line (2) is \(m_2 =\frac{ -1}{\sqrt3}\)
The acute angle i.e., \(θ\) between the two lines is given by
\(Tanθ=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\)
\(Tanθ=\left|\frac{-\sqrt3+\frac{1}{\sqrt3}}{1+(-\sqrt3)(\frac{-1}{\sqrt3})}\right|\)
\(Tanθ=\left|\frac{\frac{-3+1}{\sqrt3}}{1+1}\right|\)
\(=\left|\frac{-2}{2\times\sqrt3}\right|\)
\(Tanθ=\frac{1}{\sqrt3}\)
\(θ=30º\)
Thus, the angle between the given lines is either \( 30° \)or \(180° - 30° = 150°\).
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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