Find the equation of the straight lines that pass through the point [tex]$(2,3)[tex]$[/tex] and make an angle of [tex]$[/tex]45^{\circ}[tex]$[/tex] with the line [tex]$[/tex]x-3y=2$[/tex].



Answer :

To find the equations of the lines passing through the point \((2,3)\) and making a \(45^\circ\) angle with the line \(x - 3y = 2\), we proceed as follows:

1. Determine the slope of the given line:
The equation of the line is \(x - 3y = 2\). This can be rewritten in the slope-intercept form \(y = mx + b\).
[tex]\[ y = \frac{1}{3}x - \frac{2}{3} \][/tex]
So the slope (\(m\)) of the given line is \(\frac{1}{3}\).

2. Calculate the slopes of the new lines:
Let \(\theta\) be the angle between the given line and the lines of interest, which is \(45^\circ\). The slopes \(m_1\) and \(m_2\) of the lines that make these angles can be found using the angle addition and subtraction formula for slopes:
[tex]\[ m_1 = \frac{m + \tan\theta}{1 - m\tan\theta} \quad \text{and} \quad m_2 = \frac{m - \tan\theta}{1 + m\tan\theta} \][/tex]
Here, \(m = \frac{1}{3}\) and \(\tan 45^\circ = 1\).
[tex]\[ m_1 = \frac{\frac{1}{3} + 1}{1 - \frac{1}{3} \cdot 1} = \frac{\frac{4}{3}}{\frac{2}{3}} = 2 \][/tex]
[tex]\[ m_2 = \frac{\frac{1}{3} - 1}{1 + \frac{1}{3} \cdot 1} = \frac{\frac{-2}{3}}{\frac{4}{3}} = -\frac{1}{2} \][/tex]

3. Formulate the equations of the new lines:
Using the point-slope form of the equation of a line \(y - y_1 = m(x - x_1)\), where the point \((x_1, y_1)\) is \((2, 3)\):

- For the line with slope \(m_1 = 2\):
[tex]\[ y - 3 = 2(x - 2) \][/tex]
Simplify to:
[tex]\[ y - 3 = 2x - 4 \][/tex]
[tex]\[ y = 2x - 1 \][/tex]

- For the line with slope \(m_2 = -\frac{1}{2}\):
[tex]\[ y - 3 = -\frac{1}{2}(x - 2) \][/tex]
Simplify to:
[tex]\[ y - 3 = -\frac{1}{2}x + 1 \][/tex]
[tex]\[ y = -\frac{1}{2}x + 4 \][/tex]

Thus, the equations of the lines passing through \((2, 3)\) and making an angle of \(45^\circ\) with the line \(x - 3y = 2\) are:
[tex]\[ \boxed{y = 2x - 1 \text{ and } y = -\frac{1}{2}x + 4} \][/tex]