# Fill in the blank for the the following : Equations of the lines through the point $(3, 2)$ and making an angle of $45^{\circ}$ with the line $x-2y=3$ are _______.

$\begin {array} {1 1} (A)\;3x+y=7 ; x-3y=9 & \quad (B)\;3x-y=7; x+3y=9 \\ (C)\;3x-y=-7; x-3y=-9 & \quad (D)\;3x+y=-7 ; x+3y=-9 \end {array}$

Toolbox:
• Slope of an angle is $m = \tan \theta$
• Angle between two straight lines whose slopes are $m_1$ and $m_2$ is $\tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Step 1 :
Angle made with the line $x-2y=3$ is $45^{\circ}$
Hence $\tan 45^{\circ}=1$
(i.e) $m_1=1$
Slope of the line $x-2y=3$ is $m_2 = -\bigg( \large\frac{1}{-2} \bigg)$
$m_2 = \large\frac{1}{2}$
$\therefore \tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
$= \bigg| \large\frac{1-\Large\frac{1}{2} }{1+1 \times \Large\frac{1}{2} } \bigg|$ or $\bigg| \large\frac{-1-\Large\frac{1}{2}}{1+(-1) \bigg( \Large\frac{1}{2} \bigg)} \bigg|$
$= \bigg| \large\frac{\Large\frac{1}{2}}{\Large\frac{3}{2}} \bigg|$$= \large\frac{1}{3} = \bigg|\large\frac{- \Large\frac{3}{2}}{\Large\frac{1}{2}} \bigg|$$ = 3$
Hence the slope of the required line are 3 and $-\large\frac{1}{3}$
It passes through the point (3,2)
Hence the equations of the lines are where $m = 3$
$y-2 = 3(x-3)$
$\Rightarrow (y-2)=3(x-3)$
$\Rightarrow y-2=3x-9$
or $3x-y=7$--------(1)
where $m = -\large\frac{1}{3}$
$y-2=-\large\frac{1}{3}$$(x-3)$
$3y-6=-x+3$
$\Rightarrow x+3y=9$---------(2)
Hence the required equations of the line are
$3x-y=7$ or $x+3y=9$
answered Jul 8, 2014