confusion with part ii of this q (1 Viewer)

[amdspotter]

 

[jimmysmith560]

Would this help?

 

[amdspotter]

Would this help?

View attachment 34425

Could you send any working out if possible

 

[jimmysmith560]

Could you send any working out if possible

The working for this part is just the diagram. I can provide working for the other parts if needed.

 

[yanujw]

@amdspotter is a locus of numbers where the distance to a is the distance to b, for any point on the locus. Obviously the midpoint of a and b is included, and it follows that the perpendicular bisector of the a and b retains the same distance from a and b.

 

[CM_Tutor]

Put slightly differently...

Let and be the points such that and where is the origin.

The locus is the set of points such that the distance from to and the distance from to are equal.

The distance from to is and similarly .

It is clear that , the midpoint of , must be one such point on the locus. The diagram provided makes it clear that the origin is another point on the locus.

Now, for any point other than on the perpendicular bisector of , creates a pair of congruent triangles, which proves that and lies on the locus.

For any point that is not on the perpendicular bisector of , we can prove that either is closer to than to , or vice versa - that is closer to than to . In either case, does not belong to the locus and so the locus is all points lying on the perpendicular bisector of .

This result is easy to establish algebraically, by taking and using .