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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