Plz help me solve The locus of a point which moves in a plane such that it is equidistant from

two fixed points X and Y is

A. the perpendicular bisector of the line segment X Y

B. a line parallel to the line segment XY C. a circle with XY as diameter

D. the line perpendicular to the lin

  A. the perpendicular bisector of the line segment XY

Step-by-step explanation:

You know that one point equidistant from X and Y is the midpoint of segment XY. The third vertex of an isosceles triangle with XY as its base will also be equidistant from X and Y. That vertex will be located on the perpendicular bisector of XY.

The locus of points of interest is the perpendicular bisector of XY.

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For many geometry problems, it can be helpful to draw a diagram.

a4 - 3a2 - 1

step-by-step explanation:

step   1   :

equation at the end of step   1   :

  ((a4) -   3a2) -   1

step   2   :

trying to factor by splitting the middle term

2.1     factoring   a4-3a2-1  

the first term is,   a4   its coefficient is   1 .

the middle term is,   -3a2   its coefficient is   -3 .

the last term, "the constant", is   -1  

a

step-by-step explanation:

u take $30 minus $24. the answer/original($30)