The equation of the locus of points on the plane OXY equidistant from points A (5; 4) and B (7; -2) has the form

We need to draw up an equation of the locus of points on the plane ОXY equidistant from points with coordinates A (5; 4) and B (7; – 2).

We will solve the problem as follows:

recall the formula for finding the distance between points on a plane;
denote the points equidistant from A and B by coordinates (x; y);
write down the distance between point A and (x; y);
write down the distance between points B and (x; y);
let’s equate the distances and express one variable in terms of another.
Recall the formula for finding the distance on the plane
The formula for finding the distance between points on a plane looks like this:

AB = √ (xb – xa) ^ 2 + (yb – ya) ^ 2, where points A and B are given by coordinates A (xa, ya) and B (xb, yb).

We remembered the formula, now we can write down the distance between points A with coordinates (5; 4) and (x; y) and points B with coordinates (7; – 2) and (x; y).

Let’s compose the equation of the locus of points
We write down the distance between point A (5; 4) and (x; y):

√ ((x – 5) ^ 2 + (y – 4) ^ 2);

We write down the distance between points B (7; – 2) and (x; y):

√ ((x – 7) ^ 2 + (y + 2) ^ 2;

Since the locus of points on the plane ОXY equidistant from points A and B, we equate the obtained expressions:

√ ((x – 5) ^ 2 + (y – 4) ^ 2) = √ ((x – 7) ^ 2 + (y + 2) ^ 2;

(x – 5) ^ 2 + (y – 4) ^ 2 = (x – 7) ^ 2 + (y + 2) ^ 2;

We open the brackets, transfer all the terms to the right and give similar ones.

x ^ 2 – 10x + 25 + y ^ 2 – 8y + 16 = x ^ 2 – 14x + 49 + y ^ 2 + 4y + 4;

x ^ 2 – x ^ 2 – 10x + 14x + y ^ 2 – y ^ 2 – 8y – 4y + 25 + 16 – 49 – 4 = 0;

4x – 12y – 12 = 0;

x – 3y – 3 = 0;

x = 3y + 3;

or

y = (x – 3) / 3 = x / 3 – 1.

Answer: y = x / 3 – 1 and x = 3y + 3.