Consider $O$ as the origin and a variable straight line drawn through it to cut $a_{1}x+b_{1}y+1=0$ and $a_{2}+b_{2}y+1=0$ in $L$ and $M$ respectively. Let $N$ be a point on the variable line. Find the locus of $N$ in each of the following cases. $ON$ is (i) arithmetic mean, (ii) geometric mean and (iii) harmonic mean of $OL$ and $OM$ respectively.
Consider the parametric form of straight line. $x_{1}=r\cos\theta, y_{1}=r\sin\theta$. Also from their intersection we have the following equalities.
$$\begin{aligned}\boxed{r_{1}=\frac{-1}{a_{1}\cos\theta+b_{1}\sin\theta}}\end{aligned}\\\boxed{r_{2}=\frac{-1}{a_{2}\cos\theta+b_{2}\sin\theta}} $$
For the first case we have $2r=r_{1}+r_{2}$. Putting values of these in terms of $x_{1}$ and $y_{1}$ gives the following equation for the locus of $N$.
$$\begin{aligned}2&=-\frac{(a_{1}+a_{2})x_{1}+(b_{1}+b_{2})y_{1}}{a_{1}a_{2}x_{1}^2+b_{1}b_{2}y_{1}^2+(a_{1}b_{2}+a_{2}b_{1})x_{1}y_{1}}\\0&=2a_{1}a_{2}x_{1}^2+2b_{1}b_{2}y_{1}^2+2(a_{1}b_{2}+a_{2}b_{1})x_{1}y_{1}\color{red}{+}(a_{1}+a_{2})x_{1}\color{red}{+}(b_{1}+b_{2})y_{1}\end{aligned}$$
However the given answer has $-$ sign in place of $\color{red}{+}$. What is wrong? Thanks.