Straight LineHardBloom L1
Question
The distance of any point ( $x,y$ ) from the origin is defined as $d = max\{|x|,|y|\}$, then the distance of the common point for the family of lines $x(1 + \lambda) + \lambda y + 2 + \lambda = 0$ ( $\lambda$ being parameter) from origin is :
Options
A.1
B.2
C.$\sqrt{5}$
D.0
Solution
Sol. [Key Concept: Locate the fixed point of the family, then use the distance $d = max\{|x|,|y|\}$.]
$$(x + 2) + \lambda(x + y + 1) = 0$$
Fixed point, $P \equiv ( - 2,1)$
Distance of $P$ from origin $= 2$
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