Question

Find the equations of the lines through the point $\left( {3,2} \right)$ which makes an angle of $45^\circ $ with the line $x - 2y = 3$ are
A. $3x - y = 7$ and $x + 3y = 9$
B. $x - 3y = 7$ and $3x + y = 9$
C. $x - y = 3$ and $x + y = 2$
D. $2x + y = 7$ and $x - 2y = 9$

Answer 1

Hint: A equation of a line is given and some other conditions are given. Using the conditions you have to find two equations of two lines. To find the equation of the line, use point-slope form because a point on the lines is given and you can find the slopes easily using the formula of finding the included angle between two lines.

Formula Used:
If $\theta $ be the included angle between two lines having slopes ${m_1}$ and ${m_2}$ then $\tan \theta = \pm \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}$
The equation of a line having slope $m$ and passing through the point $\left( {{x_1},{y_1}} \right)$ is $\dfrac{{y - {y_1}}}{{x - {x_1}}} = m$

Complete step by step solution:
Write the given equation in the form $y = mx + c$ to find the slope $m$.
The given line is $x - 2y = 3$
$ \Rightarrow 2y = x - 3$
$ \Rightarrow y = \dfrac{1}{2}x - \dfrac{3}{2}$
So, slope of the line is $m = \dfrac{1}{2}$
Let the slope of the required line be $m'$
Then find the slope $m'$ using the formula of the included angle of two lines.
Here $\theta = 45^\circ $, ${m_1} = \dfrac{1}{2}$, ${m_2} = m'$
So, $\tan 45^\circ = \pm \dfrac{{\dfrac{1}{2} - m'}}{{1 + \dfrac{1}{2}m'}}$
$ \Rightarrow 1 = \pm \dfrac{{1 - 2m'}}{{2 + m'}}$
Taking positive sign, we get
$1 = \dfrac{{1 - 2m'}}{{2 + m'}}$
$ \Rightarrow 2 + m' = 1 - 2m'$
$ \Rightarrow m' + 2m' = 1 - 2$
$ \Rightarrow 3m' = - 1$
$ \Rightarrow m' = - \dfrac{1}{3}$
Taking negative sign, we get
$1 = - \dfrac{{1 - 2m'}}{{2 + m'}}$
$ \Rightarrow 2 + m' = - 1 + 2m'$
$ \Rightarrow m' - 2m' = - 1 - 2$
$ \Rightarrow - m' = - 3$
$ \Rightarrow m' = 3$
So, the slopes of the two lines are $\left( { - \dfrac{1}{3}} \right)$ and $3$.
So, the equation of the line passing through the point $\left( {3,2} \right)$ and having slope $\left( { - \dfrac{1}{3}} \right)$ is
$\dfrac{{y - 2}}{{x - 3}} = - \dfrac{1}{3}$
$ \Rightarrow 3\left( {y - 2} \right) = - 1\left( {x - 3} \right)$
$ \Rightarrow 3y - 6 = - x + 3$
$ \Rightarrow x + 3y = 9$
and the equation of the line passing through the point $\left( {3,2} \right)$ and having slope $3$ is
$\dfrac{{y - 2}}{{x - 3}} = 3$
$ \Rightarrow \left( {y - 2} \right) = 3\left( {x - 3} \right)$
$ \Rightarrow y - 2 = 3x - 9$
$ \Rightarrow 3x - y = 7$

Option ‘A’ is correct

Note: Generally, an acute angle is considered as an angle between two curves and that’s why the absolute value is taken for $\tan \theta $. But in this problem both lines are required. That’s why the $ \pm $ sign has been taken in the formula of finding the included angle between two lines. The acute angle must be chosen when only the angle is required.