On the abscissa, find a point equidistant from the points: a) A (1; 2) and B (-3; 4) b) C (1; 1) and D (3; 5).

a) The coordinates of the point on the abscissa axis are (x; 0). It is required to find x. Equation of distances from A (1; 2) and B (-3; 4) corresponds to the equation

((x – 1) ^ 2 + (0 – 2) ^ 2) ^ 0.5 = ((x – (-3)) ^ 2 + (0 – 4) ^ 2) ^ 0.5, whence

(x – 1) ^ 2 + (0 – 2) ^ 2 = ((x – (-3)) ^ 2 + (0 – 4) ^ 2

-2x + 5 = 6x + 25

4x = -20

x = – 2.5

Answer: (-2.5; 0)

b) Similarly, we obtain

((x – 1) ^ 2 + (0 – 1) ^ 2) ^ 0.5 = ((x – 3) ^ 2 + (0 – 5) ^ 2) ^ 0.5, whence

(x – 1) ^ 2 + (0 – 1) ^ 2 = ((x – 3) ^ 2 + (0 – 5) ^ 2

-2x + 2 = – 6x + 34

4x = 32

x = 8

Answer: (8; 0)