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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the equation of the lines through the point (3, 2) which make an angle of $45^{\circ}$ with the line $x-2y=3.$

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

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Toolbox:
  • Slope $m = \tan \theta$
  • Angle between two lines whose slopes are $m_1$ and $m_2$ is $\tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Let the slope of the required line be $m_1$.
Equation of the given line is $x-2y=3$
$ \qquad y = \large\frac{1}{2}$$x- \large\frac{3}{2}$
which is of the form $y=mx+c$
Hence the slope of the given line is $m_2 = \large\frac{1}{2}$ given angle between the above line and the required line is $45^{\circ}$
Hence $\tan 45^{\circ} = \large\frac{|m_1-m_2|}{1+m_1m_2}$
But $\tan 45 = 1$
$ \therefore 1 = \large\frac{|\Large\frac{1}{2}-m_1|}{1+\Large\frac{m_1}{2}}$
$ \Rightarrow 1 = \large\frac{\bigg|\bigg( \Large\frac{1-2m_1}{2} \bigg) \bigg|}{\Large\frac{2+m_1}{2}}$
$ \Rightarrow 1= \pm \bigg( \large\frac{1-2m_1}{2+m_1} \bigg)$
(i.e.), $1= \large\frac{1-2m_1}{2+m_1}$$ or 1= \large\frac{-(1-2m_1)}{2+m_1}$
If $1= \large\frac{1-2m_1}{2+m_1}$
then $2+m_1=1-2m_1$
$ \Rightarrow m_1= -\large\frac{1}{m}$
If $1= \large\frac{-(1-2m_1)}{2+m_1}$
(i.e.,) $2+m_1=-1+2m_1$
$ \Rightarrow m_1=3$
When $m_1=3$
The equation of the line passing through (3, 2) and slope 3 is
$y-2=3(x-3)$
$ \Rightarrow y-2=3x-9$
$3x-y=7$
When $m_1= -\large\frac{1}{3}$
The equation of the line passing through (3,2) and having a slope $ -\large\frac{1}{3}$ is
$ y-2=-\large\frac{1}{3}$$(x-3)$
$3y-6=-x+3$
$x+3y=9$
answered May 20, 2014 by thanvigandhi_1
 

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