Question

Prove that the three straight lines whose equations are $15x-18y+1=0,12x+10y-3=0$, and $6x+66y-11=0$ all meet in a point.
Show also that the third line bisects the angle between the other two.

Answer 1

$15x-18y+1=\mathrm{0.......}\left(i\right)12x+10y-3=\mathrm{0..........}\left(ii\right)12x=3-10yx=\frac{3-10y}{12}$

Substituting $x$ in $\left(i\right)$, we get

$15\left(\frac{3-10y}{12}\right)-18y+1=0\frac{15-50y-72y+4}{4}=019-122y=0y=\frac{19}{122}x=\frac{3-10y}{12}=\frac{3-10\left(\frac{19}{122}\right)}{12}x=\frac{366-190}{\left(122\right)12}=\frac{22}{183}$

So the point of intersection is $P(\frac{22}{183},\frac{19}{122})$

$6x+66y-11=\mathrm{0......}\left(iii\right)$

Substituting $P$ in $\left(iii\right)$

$6\times \frac{22}{183}+66\times \frac{19}{122}-11=0\frac{44}{61}+\frac{627}{61}-11=0\frac{671}{61}-11=011-11=00=0$

$P$ also satisfies the equation of third line, so they all meet in a point .

Equation of angle bisector between $15x-18y+1=0$ and $12x+10y-3=0$ is

$\frac{15x-18y+1}{\sqrt{{\left(15\right)}^2+{\left(18\right)}^2}}=\pm \frac{12x+10y-3}{\sqrt{{\left(12\right)}^2+{\left(10\right)}^2}}\frac{15x-18y+1}{\sqrt{549}}=\pm \frac{12x+10y-3}{\sqrt{244}}\frac{15x-18y+1}{3\sqrt{61}}=\pm \frac{12x+10y-3}{2\sqrt{61}}2(15x-18y+1)=\pm 3(12x+10y-3)30x-36y+2=\pm (36x+30y-3)\Rightarrow 30x-36y+2=36x+30y-96x+66y-11=0$

which is same as $\left(ii\right)$

Hence, third line bisects the angle between the other two.