Points A (1; 3), B (2; 6) and C (5; 7) are the vertices of the parallelogram ABCD. Find the coordinates of the vertex D

Let’s use the equality of sides AB and CD. Let’s find the length of this side:

| AB | = sqrt ((2 – 1) 2 + (6 – 3) 2) = sqrt (10).

The same, only for sides BC and AD:

| BC | = sqrt ((5 – 2) 2 + (7 – 6) 2) = sqrt (10).

Our parallelogram turned out to be a rhombus, since the adjacent sides are equal. Then let’s equate the AD and CD sides:

sqrt ((x – 1) 2 + (y – 3) 2) = sqrt ((x – 5) 2 + (y – 7) 2);

(x – 1) 2 + (y – 3) 2 = (x – 5) 2 + (y – 7) 2;

(x – 1) 2 – (x – 5) 2 = (y – 7) 2 – (y – 3) 2;

(x – 1 – x + 5) * (x – 1 + x – 5) = (y – 7 – y + 3) * (y – 7 + y – 3);

4 * (2 * x – 6) = – 4 * (2 * y – 10);

2 * (x – 3) = – 2 * (y – 5);

x – 3 = 5 – y;

x = 8 – y.

Substitute the expression for x and find y:

(8 – y – 1) 2 + (y – 3) 2 = 10;

49 – 14 * y + y2 + y2 – 6 * y + 9 – 10 = 0.

2 * y2 – 20 * y – 48 = 0;

y2 – 10 * y – 24 = 0;

y1 = 12; x1 = 8 – 12 = – 4;

y2 = – 2; x2 = 8 + 2 = 10.

Answer: (- 4; 12) and (10; – 2).