Question

Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line  \(x -2y = 3.\)

Answer 1

Let the slope of the required line be  \(m_1\). 
The given line can be written as \(y =\frac{ 1}{2} x – \frac{3}{2} ,\) which is of the form \(y = mx + c\)

∴ Slope of the given line = \(m_2 = \frac{1}{2}\)

It is given that the angle between the required line and line  \( x - 2y = 3 \) is 45°. 
We know that if θisthe acute angle between lines \(l_1\) and \(l_2\) with slopes \(m_1\) and  \( m_2\) respectively, then

\(tanθ=\left|\frac{m_2-m_1}{1+m_1m_2}\right|\)

\(∴  tan45º=\left|\frac{m_2-m_1}{1+m_1m_2}\right|\)

\(⇒1=\left|\frac{\frac{1}{2}-m_1}{1+\frac{m_1}{2}}\right|\)

\(⇒1=\left|\frac{\left(\frac{1-2m_1}{2}\right)}{\frac{2+m_1}{2}}\right|\)

\(⇒1=\left|\frac{1-2m_1}{2+m_1}\right|\)

\(⇒1=±\left(\frac{1-2m_1}{2+m_1}\right)\)

\(⇒1=\left(\frac{1-2m_1}{2+m_1}\right)\) or \(1=-\left(\frac{1-2m_1}{2+m_1}\right)\)

\(⇒2+m_1=1-2m_1\) or \(2+m_1=-1+2m_1\)

\(⇒m_1=\frac{-1}{3}\) or  \( m_1=3\)

Case I:

\(m_1 = 3 \)
The equation of the line passing through (3, 2) and having a slope of 3 is:
\(y -2 = 3 (x - 3) \)
\(y - 2 = 3x - 9 \)
\(3x - y = 7\)

Case II:

\(m_1 =\frac{-1}{3}\)
The equation of the line passing through (3, 2) and having a slope of \(\frac{-1}{3} \) is:
\(y – 2 = – \frac{1}{3} (x – 3)\)

\(3y – 6 = – x + 3\)
\(x + 3y = 9\)

Thus, the equations of the lines are \(3x - y = 7 \) and \(x + 3y = 9.\)