Question

Equation of a bisector of the angle between the lines $y-b=\frac{2m}{1-m^2}(x-a)$ and $y-b=\frac{2m{}^{\prime }}{1-m{}^{\prime 2}}(x-a)$ is

• A. $(y-b)(m+m{}^{\prime })+(x-a)(1-mm{}^{\prime })=0$
• B. $(y-b)(1-mm{}^{\prime })+(x-a)(m+m{}^{\prime })=0$
• C. $(x-a)(m+m{}^{\prime })+(y-b)(1-mm{}^{\prime })=0$
• D. $(x-a)(m+m{}^{\prime })-(y-b)(1-mm{}^{\prime })=0$

Answer 1

Answer: (A), (D)

The correct options are
A $(y-b)(m+m{}^{\prime })+(x-a)(1-mm{}^{\prime })=0$
D $(x-a)(m+m{}^{\prime })-(y-b)(1-mm{}^{\prime })=0$

Equations of the bisectors are given by
$\frac{\frac{2m}{1-m^2}(x-a)-(y-b)}{\sqrt{{\left(\frac{2m}{1-m^2}\right)}^2+1}}=\pm \frac{\frac{2m{}^{\prime }}{1-m{}^{\prime 2}}(x-a)-(y-b)}{\sqrt{{\left(\frac{2m{}^{\prime }}{1-m{}^{\prime 2}}\right)}^2+1}}=$

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

Taking the positive sign, we get

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

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

Similarly, by taking the negative sign, we get

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