Find the coordinates of a point belonging to the ordinate axis and equidistant from points C (3, 2) and D (1; -6)

We denote by y the ordinate of the desired point, and the desired point itself by A.

According to the condition of the problem, this point belongs to the ordinate axis, therefore, the abscissa of this point is 0 and this point has coordinates (0; y).

According to the condition of the problem, point A must be equidistant from points C (3; 2) and D (1; -6), therefore, the equality must be fulfilled:

| AC | = | AD |.

Applying the formula for the distance between two points A and B on the coordinate plane, we can write:

| AC | = √ ((3 – 0) ^ 2 + (2 – y) ^ 2) = √ (3 ^ 2 + (2 – y) ^ 2) = √ (9 + 4 – 4y + y ^ 2) = √ ( y ^ 2 – 4y + 13);

| AD | = √ ((1 – 0) ^ 2 + (-6 – y) ^ 2) = √ (1 ^ 2 + (6 + y) ^ 2) = √ (1 + 36 + 12y + y ^ 2 ) = √ (y ^ 2 + 12y + 37)

and we can make the following equation:

√ (y ^ 2 – 4y + 13) = (y ^ 2 + 12y + 37),

solving which, we get:

y ^ 2 – 4y + 13 = y ^ 2 + 12y + 37;

-4y + 13 = 12y + 37;

12y + 4y = 13 – 37;

16y = 24;

y = 24/16 = 3/2 = 1.5.

Answer: the coordinates of the desired point (0; 1.5).