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

- 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$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...