The locus of the orthocentre of the triangle | vedclass

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(D) Let the lines be $L_1: (1+p)x - py + p(1+p) = 0$,$L_2: (1+q)x - qy + q(1+q) = 0$,and $L_3: y = 0$.The vertices of the triangle are found by solvin

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a straight line

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(D) Let the lines be $L_1: (1+p)x - py + p(1+p) = 0$,$L_2: (1+q)x - qy + q(1+q) = 0$,and $L_3: y = 0$.The vertices of the triangle are found by solving the equations in pairs:Intersection of $L_1$ and $L_3$ $(y=0)$: $(1+p)x + p(1+p) = 0 \Rightarrow x = -p$. So,$A = (-p, 0)$.Intersection of $L_2$ and $L_3$ $(y=0)$: $(1+q)x + q(1+q) = 0 \Rightarrow x = -q$. So,$B = (-q, 0)$.Intersection of $L_1$ and $L_2$: Subtracting the equations gives $(p-q)x - (p-q)y + (p^2+p - q^2-q) = 0$. Since $p 
eq q$,we divide by $(p-q)$ to get $x - y + (p+q+1) = 0$. Solving this with $L_1$ gives $C = (pq, (1+p)(1+q))$.The altitude from $C$ to $AB$ (the $x$-axis) is the vertical line $x = pq$.The slope of $AC$ is $m_{AC} = \frac{(1+p)(1+q) - 0}{pq - (-p)} = \frac{(1+p)(1+q)}{p(q+1)} = \frac{1+p}{p}$.The altitude from $B(-q, 0)$ to $AC$ has slope $m = -\frac{p}{1+p}$.Equation: $y - 0 = -\frac{p}{1+p}(x + q)$ $\Rightarrow (1+p)y = -px - pq$ $\Rightarrow px + (1+p)y + pq = 0$.Substitute $x = pq$ into this equation: $p(pq) + (1+p)y + pq = 0$ $\Rightarrow p^2q + pq + (1+p)y = 0$ $\Rightarrow pq(p+1) + (1+p)y = 0$.Since $p 
eq -1$,we get $y = -pq$.Thus,the orthocentre $(h, k)$ is $(pq, -pq)$.Since $h = pq$ and $k = -pq$,we have $k = -h$,which represents the line $y = -x$.

The locus of the orthocentre of the triangle formed by the lines $(1+p) x-p y+p(1+p)=0$,$(1+q) x-q y+q(1+q)=0$,and $y=0$,where $p \neq q$,is

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