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 [2009]

Question

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 [2009]

Smart Achievers · Accepted Answer

(4)

$The triangle is formed by the lines$

$A B : (1 + p) x - p y + p (1 + p) = 0$

$A C : (1 + q) x - q y + q (1 + q) = 0$

$B C : y = 0$

$So that the vertices of ∆ A B C are$

$A (p q, (p + 1) (q + 1)), B (- p, 0) and C (- q, 0)$

$Let H (h, k) be the orthocentre of ∆ A B C . Then as$ $A H ⊥ B C$ $and passes through A (p q, (p + 1) (q + 1))$

$∴$ $Equation of A H is x = p q$

$∴$ $h = p q$ ...(i)

$∵$ $B H is perpendicular to A C$

$∴$ $m_{1} m_{2} = - 1$ $\Rightarrow \frac{k - 0}{h + p} \times \frac{1 + q}{q} = - 1$

$\Rightarrow \frac{k}{p q + p} \times \frac{1 + q}{q} = - 1 [using (i)]$

$∴$ $k = - p q$ ...(ii)

$From (i) and (ii), we observe that$ $h + k = 0$

$∴$ $Locus of (h, k) is x + y = 0,$ $which is a straight line.$

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