• ask the user for the coordinates x1, y1, x2, y2, x3, and y3 of the three vertices, and
• compute the three sides s1, s2, and s3, the perimeter p, the half-perimeter s, and the area a of the triangle.

• to compute the sides, use Pythagoras theorem:
  s1 = √(x2 − x3)² + (y2 − y3)²;
  s2 = √(x1 − x3)² + (y1 − y3)²;
  s3 = √(x1 − x2)² + (y1 − y2)²;
• perimeter is the sum of all three sides: p = s1 + s2 + s3;
• the half-perimeter is literally the half of the perimeter: s = 0.5 * p;
• finally, the area can be computed by using Heron's formula:
  a = √s * (s − s1) * (s − s2) * (s − s3).

How to compute square root. Use Math.sqrt(x) to compute the square root of x. For example, the instruction

double a = Math.sqrt(4.0);

To use Math.sqrt, you probably need to import java.lang.Math

Example. Test your program on the example of the following triangle:

• its first vertex has coordinates x1 = 0.0 and y1 = 0.0;
• its second vertex has coordinates x2 = 4.0 and y2 = 0.0;
• its third vertex has coordinates x3 = 0.0 and y3 = 3.0.

For this triangle, the above formulas lead to

• s1 = √(4.0 − 0.0)² + (0.0 − 3.0)² = √4.0² + (−3.0)² = √16.0 + 9.0 = √25.0 = 5.0;
• s2 = √(0.0 − 0.0)² + (0.0 − 3.0)² = √9.0 = 3.0;
• s3 = √(0.0 − 4.0)² + (0.0 − 0.0)² = √16.0 = 4.0.

Here,

• p = 5.0 + 4.0 + 3.0 = 12.0,
• s = 0.5 * 12.0 = 6.0, and
• a = √6.0 * (6.0 − 5.0) * (6.0 − 3.0) * (6.0 − 4.0) = √6.0 * 1.0 * 3.0 * 2.0 = √36.0 = 6.0.

When it is due. The program is due at the beginning of the first lab section on the week of September 15, i.e.:

• on Monday September 15 for those who attend Monday-Wednesday labs, and
• on Tuesday September 16 for those who attend Tuesday-Thursday labs.

How to submit: Your TAs will instruct you how to submit the programs.