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Find the co-ordinates of the orthocentre of the triangle formed by the lines $\mathrm{y=0}$ , $(1+\mathrm{t}) \mathrm{x}-\mathrm{ty}+\mathrm{t}(1+\mathrm{t})=0$ and $(1+\mathrm{u}) \mathrm{x}-\mathrm{uy}+\mathrm{u}(1+\mathrm{u})=0(\mathrm{t} \neq \mathrm{u})$ .

Option: 1
$\mathrm{(tu,tu)}$

Option: 2
$\mathrm{(-tu,-tu)}$

Option: 3
$\mathrm{(-t+u,tu)}$

Option: 4
$\mathrm{(tu,-tu)}$

Answers (1)

best_answer

Let equations of $\mathrm{\mathrm{BC}, \mathrm{CA} \: and\: \mathrm{AB}}$ are

$\mathrm{\mathrm{y}=0,(1+\mathrm{t}) \mathrm{x}-\mathrm{ty}+\mathrm{t}(1+\mathrm{t})=0\: and \: (1+u) x-u y+u(1+u)=0}$ respectively.

Let $\mathrm{O}(\mathrm{h}, \mathrm{k})$ be the orthocentre then

$\mathrm{Slope\: of \: \mathrm{OB} \times \: Slope\: of\: \mathrm{AC}=-1}$

$\frac{\mathrm{k}-0}{\mathrm{~h}+\mathrm{u}} \times \frac{(1+\mathrm{t})}{\mathrm{t}}=-1$

$\Rightarrow \quad \mathrm{k}(1+\mathrm{t})=-\mathrm{th}-\mathrm{tu}$

$\therefore \quad \mathrm{k}=\frac{-(\mathrm{th}+\mathrm{tu})}{(1+\mathrm{t})}$

$\Rightarrow \quad \mathrm{k}+\mathrm{tk}=-\mathrm{th}-\mathrm{tu}$ ...........(i)

and $\mathrm{slope \: of\: \mathrm{OC} \times \: slope \: of \: \mathrm{AB}=-1}$

$\frac{\mathrm{k}-0}{(\mathrm{~h}+\mathrm{t})} \times \frac{(1+\mathrm{u})}{\mathrm{u}}=-1$

$\therefore \quad \mathrm{k}=\frac{-\mathrm{u}(\mathrm{h}+\mathrm{t})}{(1+\mathrm{u})}$

$\Rightarrow \quad \mathrm{k}+\mathrm{ku}=-\mathrm{uh}-\mathrm{ut}$

Subtracting (ii) from (i), we get ..........(ii)

$\mathrm{k}(\mathrm{t}-\mathrm{u})=-\mathrm{h}(\mathrm{t}-\mathrm{u}) \Rightarrow \mathrm{h}+\mathrm{k}=0$

$\therefore$ locus of orthocentre is $\mathrm{x}+\mathrm{y}=0$

Putting $\mathrm{h}=-\mathrm{k}$ in (i)

then we get $\mathrm{k}=-\mathrm{tu}$

$\therefore \mathrm{h}=\mathrm{tu}$

then orthocentre is $\mathrm{(tu, -tu).}$

Hence option 4 is correct.

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Nehul

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