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The orthocentre of the triangle formed by the lines $y = 0, (1 + t) x - t y + t (1 + t) = 0$ and $(1 + u) x - u y$ $+ u (1 + u) = 0 (t \neq u)$ for all values of $t$ and $u$ lies on the line.

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x – y = 0

x + y = 0

x – y + 1 = 0

x + y + 1 = 0

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detailed solution

Correct option is B

The orthocentre (x, y) lies on the lines.y=−u1+u(x+t)and y=−t1+t(x+u)⇒[(1+u)−(1+t)]y+(u−t)x=0⇒(u−t)(x+y)=0⇒x+y=0 as u≠t

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The orthocentre of the triangle formed by the lines y=0,(1+t)x−ty+t(1+t)=0 and (1+u)x−uy +u(1+u)=0(t≠u) for all values of t and u lies on the line.

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The orthocentre of the triangle formed by the lines $y = 0, (1 + t) x - t y + t (1 + t) = 0$ and $(1 + u) x - u y$ $+ u (1 + u) = 0 (t \neq u)$ for all values of $t$ and $u$ lies on the line.