Is my equation for finding equidistant points correct?

[aisha]

Here is the question I think I've almost got the answer. Describe and create an equation for each locus.

a.) THe points are equidistant from points A(1,5) and B(-2,6)

I made another random point C(x,y)

AC=BC

I wrote the distance formula for both AC and BC and set them equal to each other but I am not sure how to simplify now

square root[(x+2)^2 +(y-6)^2] = square root[(x-1)^2+(y-5)^2]

I square both sides to get rid of the square root but after that i don't know what to do I got a final answer of 6x-2y+13=0 please tell me is this correct?

 

[Galileo]

Almost, just a small arithmetic error. It should be 6x-2y+14=0
Geometrically you can see it should be a line. Moreover, it should pass through the midpoint of the line joining the points A and B and be perpendicular to it.
Check that the line indeed satisfies these conditions.

 

[aisha]

Almost, just a small arithmetic error. It should be 6x-2y+14=0
Geometrically you can see it should be a line. Moreover, it should pass through the midpoint of the line joining the points A and B and be perpendicular to it.
Check that the line indeed satisfies these conditions.

I got the equation but now I have to do the describing part how do I describe the locus? How will I check that the line is indeed perpendicular to points A and B?

 

[cepheid]

It's not perpendicular to the "points" A and B. How can you be perpendicular to a point? The locus should be perpendicular to the line joining A and B, and to verify this, compare the slopes of the two lines. What can you say immediately about the slopes of two lines if they are perpendicular?

 

[aisha]

oh the slope will be a negative reciprocal then ill know if it is perpendicular thanks for pointing that out lol ur right you can't be perpendicular to a point opps lol. :rofl: