Question

Find the angle between the two lines $3x+y+12=0$ and $x+2y-1=0$. Find also the coordinates of their point of intersection and the equations of lines drawn perpendicular to them from the point $(3,-2)$.

Answer 1

$3x+y+12=\mathrm{0......}\left(i\right)$

slope of line $=m_1=-\frac{a}{b}=-3$

$x+2y-1=\mathrm{0.......}\left(ii\right)$

slope of line ${=m}_2=-\frac{a}{b}=-\frac{1}{2}$

Angle between lines i.e. $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan \theta =\left|\frac{-3-(-\frac{1}{2})}{1+(-3)(-\frac{1}{2})}\right|=\left|\frac{\frac{-6+1}{2}}{\frac{2+3}{2}}\right|=\left|\frac{\frac{-5}{2}}{\frac{5}{2}}\right|\tan \theta =1\Rightarrow \theta ={45}^{\circ }$

Now $x+2y-1=0$

$x=1-2y$

Substituting in $\left(i\right)$

$3(1-2y)+y+12=03-6y+y+12=0-5y=-15\Rightarrow y=3x=1-2\left(3\right)\Rightarrow x=-5$

So the point of intersection is $(-5,3)$

Let the slope of line perpendicular to $\left(i\right)$ be $m$

$m_{1}m=-1-3m=-1\Rightarrow m=\frac{1}{3}$

It passes through $(3,-2)$, so the equation of line is

$y+2=\frac{1}{3}(x-3)3y+6=x-3x-3y=9$

Let the slope of line perpendicualr to $\left(ii\right)$ be $m^{\prime }$

$-\frac{1}{2}{m}^{\prime }=-1\Rightarrow m^{\prime }=2$

It passes through $(3,-2)$, so the equation of line is

$y+2=2(x-3)y+2=2x-62x-y=8$