Question

Find the point of intersection and the inclination of the two lines $Ax+By=A+B$ and $A(x-y)+B(x+y)=2B$.

• A. $(1,1);{45}^0$
• B. $(1,2),{60}^0$
• C. $(2,1),{75}^0$
• D. None of these

Answer 1

Answer: (A)

$(1,1);{45}^0$

$Ax+By=A+B$ ........(i)

$A(x-y)+B(x+y)=2B$

$\Rightarrow (A+B)x+(B-A)y=2B$ .......(ii)

$\Rightarrow (A+B)x=2B-(B-A)y\Rightarrow x=\frac{2B-(B-A)y}{A+B}$

Substituting $x$ in (i), we get

$A\left(\frac{2B-(B-A)y}{A+B}\right)+By=A+B$

$\Rightarrow \frac{2AB-ABy+A^{2}y+B^{2}y+ABy}{A+B}=A+B$

$\Rightarrow 2AB-ABy+B^{2}y+A^{2}y+ABy=A^2+B^2+2AB$

$\Rightarrow (A^2+B^2)y=A^2+B^2$

$\Rightarrow y=1$

Substituting $y$ in $\left(i\right)$

$\Rightarrow Ax+B\left(1\right)=A+B\Rightarrow Ax+B=A+B\Rightarrow Ax=A\Rightarrow x=1$

So, the point of intersection is $(1,1)$.

Slope of (i), $m_1=-\frac{A}{B}$.

Slope of (ii), $m_2=-\frac{(A+B)}{B-A}=\frac{A+B}{A-B}$

$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}\Rightarrow \tan \theta =\frac{-\frac{A}{B}-\frac{A+B}{A-B}}{1-\frac{A}{B}\times \frac{A+B}{A-B}}$

$\tan \theta =-\frac{\left\{\frac{A^2+AB-AB+B^2}{B(A-B)}\right\}}{\left\{\frac{{-B}^2+AB-A^2-AB}{B(A-B)}\right\}}\tan \theta =-\frac{A^2+B^2}{{-(A}^2+B^2)}=1$

$\Rightarrow \theta ={45}^o$