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Constructions

Question
CBSEENMA9002885

Given two points A and B and a positive real number k. Find the locus of a point P such that ar(ΔPAB) = k.

Solution

Given: Two points A and B and a positive real number k.
To find: The locus of a point P such that ar(ΔPAB) = k.
Construction: Draw PM ⊥ AB.
Determination: Let PM = h
               space space space space space space space space space space space space space space space space space space space space space ar left parenthesis increment PAB right parenthesis equals straight k space space space space space space space space space space space space space space space space space space space space vertical line Given
rightwards double arrow space space space space space space space space space space space space space space space 1 half left parenthesis AB right parenthesis left parenthesis PM right parenthesis equals straight k
rightwards double arrow space space space space space space space space space space space space space space space space 1 half left parenthesis AB right parenthesis left parenthesis straight h right parenthesis equals straight k
rightwards double arrow space space space space space space space space space space space space space space space space space space space space space straight h space equals space fraction numerator 2 straight k over denominator AB end fraction

∵ Points A and B are given.
∴ AB is fixed.
Also, k being a positive real number k is fixed. ∴ h is a fixed positive real number.
∴ The locus of P is a line parallel to the line
AB at a fixed distance