By the coordinates of the vertices of the triangle ABC, find the perimeter of the triangle, the equations

By the coordinates of the vertices of the triangle ABC, find the perimeter of the triangle, the equations of the sides AB and BC if A / 2; -1 /, B / 5; 3 / and C / 5; -2 /

We have the coordinates of the vertices of three points:

A (2; -1), B (5; 3), C (5; -2).

Find the lengths of the segments:

| AB | = ((5 – 2) ^ 2 + (3 + 1) ^ 2) ^ (1/2) = (9 + 16) ^ (1/2) = 5;

| AC | = ((5 – 2) ^ 2 + (-2 + 1) ^ 2) ^ (1/2) = (9 + 1) ^ (1/2) = 10 ^ (1/2);

| BC | = ((5 – 5) ^ 2 + (-2 – 3) ^ 2) ^ (1/2) = 5.

The perimeter of the triangle is:

P = 5 + 5 + 10 ^ (1/2) = 10 + 10 ^ (1/2) cm.

1) A (2; -1), B (5; 3):

-1 = 2 * k + b;

3 = 5 * k + b;

Subtract the first from the second equation:

3 * k = 4;

k = 4/3;

b = 3 – 5 * k = -11/3;

y = 4/3 * x – 11/3 is the equation of the straight line AB.

B (5; 3), C (5; -2).

x = 5 is the equation of the line BC.

3 = 5 * k + b;

-2 = 5 * k + b;

5 = 0 – wrong.