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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- 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)$

$3y-6=-x+3$

$x+3y=9$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...