The locus of the orthocentre of the triangle formed by the lines...

Given, lines are (1+p) x-p y+p(1+p)=0 and $(1+q) x-q y+q(1+q)=0$ On solving Eqs. (i) and (ii), we get C\{p q,(1+p)(1+q)\} $\therefore$ Equation of altitude $C M$ passing through $C and perpendicular to$A B$is x=p q$\because$Slope of line (ii) is \left(\frac{1+q}{q}\right)$. $\therefore$ Slope of altitude $B N$ (as shown in figure) is $\frac{-q}{1+q}$. $<img src="/api/proxy-image?url=https$ \\ $\therefore$ Equation of $B N$ is $y-0=\frac{-q}{1+q}(x+p)$ \Rightarrow \quad y=$\frac{-q}{(1+q)}($x+p) Let orthocentre of triangle be $H(h, k)$, which is the point of intersection of Eqs. (iii) and (iv) On solving Eqs. (iii) and (iv), we get $x=p q$ and $y=-p q \Rightarrow h=p q$ and $k=-p q \Rightarrow h+k=0\therefore$ Locus of $H(h, k)$ is $x+y=0$.