Two lines passing through the point (2, 3) intersects each other at an angle of \(60º .\) If slope of one line is 2, find equation of the other line.
Solution and Explanation
It is given that the slope of the first line, \(m_1 = 2. \)
Let the slope of the other line be \(m_2\).
The angle between the two lines is \(60°.\)
\(∴tan60º=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\)
\(⇒\sqrt3=\left|\frac{2-m_2}{1+2m_2}\right|\)
\(⇒\sqrt3=±\left(\frac{2-m_2}{1+2m_2}\right)\)
\(⇒\sqrt3=\left(\frac{2-m_2}{1+2m_2}\right) or \sqrt3=-\left(\frac{2-m_2}{1+2m_2}\right)\)
\(⇒\sqrt3(1+2m_2)=2-m_2 \space or \sqrt3(1+2m_2)=-(2-m_2)\)
\(⇒\sqrt3+2\sqrt3m_2+m_2=2 \space or \sqrt3+2\sqrt3m_2-m_2=-2\)
\(⇒m_2=\frac{2-\sqrt3}{\left(2\sqrt3+1\right)}\space or\space m_2=-\frac{\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}\)
Case I: \(m_2=\frac{2-\sqrt3}{\left(2\sqrt3+1\right)}\)
The equation of the line passing through point (2, 3) and having a slope of\(\frac{\left(2-\sqrt3\right)}{\left(2\sqrt3+1\right)}\) is
\((y-3)=\frac{2-\sqrt3}{2\sqrt3+1}(x-2)\)
\((2\sqrt3+1)(y-3)=(2-\sqrt3)x-2(2-\sqrt3)\)
\((\sqrt3-2)x+(2\sqrt3+1)y=-4+2\sqrt3+6\sqrt3+3\)
\((\sqrt3-2)x+(2\sqrt3+1)y=-1+8\sqrt3\)
In this case, the equation of the other line is \((\sqrt3-2)x+(2\sqrt3+1)y=-1+8\sqrt3\)
Case II: \(m_2=\frac{-\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}\)
The equation of the line passing through point (2, 3) and having a slope of\( \frac{-\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}\) is
\((y-3)=\frac{-\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}\left(x-2\right)\)
\((2\sqrt3-1)y-3(2\sqrt3-1)=-(2+\sqrt3)x+2(2+\sqrt3)\)
\((2\sqrt3-1)y+(2+\sqrt3)x=4+2\sqrt3+6\sqrt3-3\)
\((2+\sqrt3)x+(2\sqrt3-1)y=1+8\sqrt3\)
In this case, the equation of the other line is\( (2+\sqrt3)x+(2\sqrt3-1)y=1+8\sqrt3\)
Thus, the required equation of the other line is\( (\sqrt3-2)x+(2\sqrt3+1)y\)\(=-1+8\sqrt3\) or \( (2+\sqrt3)x+(2\sqrt3-1)y=1+8\sqrt3\)
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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