given two points A and B and a positive real number X. find the locus of a point P such that area of triangle PAB is equal to X
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Construct PM perpendicular to AB
Let PM=h
ar(triangle PAB) =x(given)
So, 1/2(PM)(AB) =x
1/2(h)(AB) =x
h=2x/AB
As points A and B are given, so AB is fixed.
Also x being a positive real number, x is fixed. Therefore h is a fixed positive real number.
Therefore the locus of P is a line parallel to the line AB at a fixed distance 2x/AB on either side of it.
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Answer:
Construct PM perpendicular to AB
Let PM=h
ar(triangle PAB) =x(given)
So, 1/2(PM)(AB) =x
1/2(h)(AB) =x
h=2x/AB
As points A and B are given, so AB is fixed.
Also x being a positive real number, x is fixed. Therefore h is a fixed positive real number.
Therefore the locus of P is a line parallel to the line AB at a fixed distance 2x/AB on either side of it.
Step-by-step explanation:
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