Straight LineHardBloom L3
Question
$ABC$ is a variable triangle such that $A$ is $(1,2)B$ and $C$ lie on $y = x + \lambda$ (where $\lambda$ is variable), then locus of the orthocentre of triangle $ABC$ is :
Options
A.$(x - 1)^{2} + y^{2} = 4$
B.$x + y = 3$
C.$2x - y = 0$
D.$x + 2y = 0$
Solution
Sol. [Key Concept: Vertex fixed and the opposite side sliding parallel ⇒ the orthocentre moves on the line through that vertex perpendicular to the sliding family.]
Equation of $BC$ varies
∴ Orthocenter will always lie on line perpendicular to $y = x + \lambda$ passing through $A(1,2)$.
⇒ Locus of orthocenter ' $H$ ' is $x + y = 3$.
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