Mathematics
A and B are two fixed points in a plane. Then, the locus of a point P which moves in such a way that PA2 + PB2 = AB2, is:
a square with AB as one of its sides.
a rectangle with AB as one of its sides.
a rhombus with AB as one of its diagonal.
the circumference of a circle with AB as its diameter.
Answer
The condition PA2 + PB2 = AB2 is exactly the Pythagorean relation.
This holds only when ∠APB = 90°.
All points that subtend a right angle at AB lie on a circle with AB as its diameter.
Therefore, the locus of point P is the circle having AB as its diameter.
Hence, option 4 is the correct option.
Related Questions
The locus of a point equidistant from two concentric circles is:
a circle concentric with the given circles and inside the smaller circle.
a circle concentric with the given circles and outside the larger circle.
a circle concentric with the given circles and midway between them.
the common centre of the two circles.
The locus of a point which is equidistant from a given circle consists of:
a pair of circles concentric with the given circle.
a circle concentric with the given circle and inside it.
a circle concentric with the given circle and outside it.
a pair of lines parallel to each other on either side of the centre.
The locus of the tip of the pendulum of a clock is:
a circle
a chord of a circle
an arc of a circle
a diameter of a circle
Assertion (A): In the figure, D is the mid-point of BC and AD ⟂ BC. If E lies on AD, then BE = CE.
Reason (R): Every point on the perpendicular bisector of a line segment is equidistant from its end points.
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
