Q5 of 28 Page 107

Prove that the locus of the centres of circles passing through points A and B is the perpendicular bisector of line segment AB.


Given that a circle with centre I and passing through points A and B


To prove: IC is the perpendicular bisector of AB


Proof: Here, IA=IB (radius)


So, ∆IAB is isosceles triangle.


So, ∠IAC=∠IBC………………..(1)


As we know that the altitude IC on AB is perpendicular bisector of AB.


Hence, ∠ICA=∠ICB=90°


And AC=BC


As ∆ICA and ∆ICB are congruent triangles.


Proved.


More from this chapter

All 28 →
3

What will the locus of a point equidistant from three non-collinear points A, B and C? Explain the reason of your answer.

 4

What will be the locus of a point equidistant from three collinear points? Explain the reason of your answer.

 6

In the given figure on the common base BC there are two isosceles triangles ΔPBC and ΔQBC on the opposite sides of BC. Prove that the line joining the points P and Q bisects line BC at right angles.


 7

In the given figure two isosceles triangle PQR and SQR lie on the same side of the common base QR. Prove that line SP is the perpendicular bisector of line QR.