logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Straight Lines
0 votes

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 _______.

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

Can you answer this question?
 
 

1 Answer

0 votes
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 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...