Mathematics - Point Question with Solution | TestHub

Given, lines are $$(1+p)x-py+p(1+p)=0$$ and $(1+q)x-qy+q(1+q)=0$ On solving Eqs. (i) and (ii), we get $$C\{pq,(1+p)(1+q)\}$$ $\therefore$ Equation of altitude $CM$ passing through $C$ and perpendicular to $AB$ is $$x=pq$$ $\because$ Slope of line (ii) is $\left(\frac{1+q}{q}\right)$. $\therefore$ Slope of altitude $BN$ (as shown in figure) is $\frac{-q}{1+q}$.
$\therefore$ Equation of $BN$ is $y-0=\frac{-q}{1+q}(x+p)$ $$\Rightarrow y=\frac{-q}{(1+q)}(x+p)$$ 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 $x=pq$ and $y=-pq\Rightarrow h=pq$ and $k=-pq\Rightarrow h+k=0$ $\therefore$ Locus of $H(h,k)$ is $x+y=0$.