In the x y -plane, the set of points equidistant from the points ( − 3 , 1 ) and ( 2 , 1 ) is

One way to write the equation for the line that meets this requirement is to make use of the distance formula. The distance (d) from (x, y) to (a, b) is found from the Pythagorean theorem to be

... d = √((x-a)² +(y-b)²)

This can be easier to use here if we square it.

... d² = (x -a)² + (y -b)²

Since the distance of (x, y) from the given points is the same, we can write

... (x +3)² +(y -1)² = d² = (x -2)² +(y -1)²

We can subtract the y-terms from both sides and expand the x-terms to get

... x² +6x +9 = x² -4x +4

Subtracting x²-4x+9 gives

... 10x = -5

... x = -1/2 . . . . . . The equation of all points equidistant from the given ones.

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Perhaps you noticed that the two points are on the horizontal line y=1. The midpoint between them is

... ((-3, 1) + (2, 1))/2 = (-1, 2)/2 = (-1/2, 1)

Then the points equidistant from those given will be the vertical line x=-1/2 (as above).

i think its a and d

step-by-step explanation: