Question

Find the equations to the straight lines bisecting the angles between the following pairs of straight lines, placing first the bisector of the angle in which the origin lies.
$y-b=\frac{2m}{1-m^2}(x-a)$ and $y-b=\frac{2m^{\prime }}{1-m^{\prime 2}}(x-a)$

Answer 1

Equation of lines are

$y-b=\frac{2m}{1-m^2}(x-a)2mx-(1-m^2)y+\left\{b\right(1-m^2)-2am\}=0......\left(i\right)y-b=\frac{2m^{\prime }}{1-m^{\prime 2}}(x-a)2m^{\prime }x-(1-{m^{\prime }}^2)y+\left\{b\right(1-{m^{\prime }}^2)-2a{m}^{\prime }\}......\left(ii\right)$
$\frac{2mx-(1-m^2)y+\left\{b\right(1-m^2)-2am\}}{\sqrt{{\left(2m\right)}^2+{(1-m^2)}^2}}=\pm \frac{2m^{\prime }x-(1-{m^{\prime }}^2)y+\left\{b\right(1-{m^{\prime }}^2)-2a{m}^{\prime }\}}{\sqrt{{\left(2m^{\prime }\right)}^2+{(1-{m^{\prime }}^2)}^2}}\frac{2mx-(1-m^2)y+\left\{b\right(1-m^2)-2am\}}{1+m^2}=\pm \frac{2m^{\prime }x-(1-{m^{\prime }}^2)y+\left\{b\right(1-{m^{\prime }}^2)-2a{m}^{\prime }\}}{1+m^{\prime 2}}$

Taking the positive sign we get

$2x(m+{m^{\prime }}^2-m^{\prime }-m^{\prime }{m}^2)-y\left\{\right(1-m^2\left)\right(1+m^{\prime 2})-(1-m^{\prime 2}\left)\right(1+m^2\left)\right\}+$

$b\left\{\right(1-m^2\left)\right(1+m^{\prime 2})-(1-m^{\prime 2}\left)\right(1+m^2\left)\right\}-2a(m+{m^{\prime }}^2-m^{\prime }-m^{\prime }{m}^2))=0$

$2x(m-m^{\prime })(1-m{m}^{\prime })-2y(m^{\prime }-m)(m^{\prime }+m)+2b(m^{\prime }-m)(m^{\prime }+m)-2a(m-m^{\prime })(1-m{m}^{\prime })=0(1-m{m}^{\prime })(x-a)+(m+m^{\prime })(y-b)=0$

Similarly taking the negative sign we get

$(y-b)(1-m{m}^{\prime })-(x-a)(m+m^{\prime })=0$

Hence proved