IIT JEE 1982 Maths - 'Adapted' - FIB to Single-Digit Integer Q8

Geometry Level 3

If A A and B B are points in the plane such that P A / P B = k > 0 PA/PB=k>0 ( k k is some constant) for all P P on a given circle, then the value of k k cannot be equal to ________ . \text{\_\_\_\_\_\_\_\_}.

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The answer is 1.

Shubhamkar Ayare by Shubhamkar Ayare , 22, India

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1 solution

This is a proof which shows that for k=1, The locus is not a circle but a straight line.

If k=1, P A P B = 1 \frac{PA}{PB}=1 or P A = P B PA=PB . By converse of perpendicular bisector theorem, P lies on the perpendicular bisector of P. Thus, the locus is a straight line instead of a circle.

Converse of perpendicular bisector theorem follows from the fact that the median, altitude and angle bisector are the same for the angle formed by the 2 equal sides for an isoceles triangle.

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Ajinkya Shivashankar - 4 years, 4 months ago

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