The locus of the orthocentre of the triangle | vedclass

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Question

The correct option is $D y=-x$

Let $L_1:(1+p) x-p y+p(1+p)=0$,

$L_2:(1+q) x-q y+q(1+q)=0$ and

$L_3: y=0$

The coordinates of vertices are

Vedclass · Accepted Answer

a straight line

Vedclass · Answer

The correct option is $D y=-x$

Let $L_1:(1+p) x-p y+p(1+p)=0$,

$L_2:(1+q) x-q y+q(1+q)=0$ and

$L_3: y=0$

The coordinates of vertices are

$A(-p, 0), B(-q, 0), C(p q,(1+p)(1+q))$

The equation of the altitude through $C$ is $x=p q \cdots(1)$

The equation of the altitude through $B$ is $px +(1+ p ) y +\lambda=0$

Putting $B(-q, 0)$, we get

$-p q+\lambda=0 \Rightarrow \lambda=p q$

Now, equation of altitude through $B$

$p x+(1+p) y+p q=0 \cdots$

Solving equation $(1)$ and $(2)$, we get

$p^2 q+(1+p) y+p q=0$

$\Rightarrow y=-p q$

Let $(h, k)$ be the coordinates of the orthocentre

$h=p q, k=-p q$

$\Rightarrow k=-h$

Hence, the locus of orthocentre is $y=-x$.

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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