Types and Expressions I: Computing Slopes

The Problem

Write a program that reads values x1,y1,x2,y2 and prints out the slope of the line passing through (x1,y1) and (x2,y2). The slope is given by:
    slope = (y1 - y2) / (x1 - x2)
    
For the purists among you, you may assume that the line is not vertical.

A Solution

We already know the basic structure here, just our usual "main" skeleton.
int main()
{

  return 0;
}
Let's add in comments outlining the basic process for solving this problem.
int main()
{
  // Read first point

  // Read second point

  // Compute the slope
  
  // Print out slope

  return 0;
}
	  
Reading in a point means reading in a value for x1 and y1. First let's read in x1. First we declare variable for the value, then we ask the user for the value, then we read the value!
  // Read first point
  double x1;  
  cout << "Enter x1: ";
  cin >> x1;
  double y1;  
  cout << "Enter y1: ";
  cin >> y1;
This is easily extended for the other values.
int main()
{
  // Read first point
  double x1;  
  cout << "Enter x1: ";
  cin >> x1;
  double y1;  
  cout << "Enter y1: ";
  cin >> y1;

  // Read second point
  double x2;
  cout << "Enter x2: ";
  cin >> x2;
  double y2;
  cout << "Enter y2: ";
  cin >> y2;

  // Compute the slope

  // Print out slope

  return 0;
}
To compute the slope, we first declare a variable to contain the slope, and then we compute its value.
  // Compute the slope
  double slope;
  slope = (y1 - y2) / (x1 - x2);
  
All we have to do now is print out the answer, which is old hat.
  // Print out slope
  cout << "The slope is: ";
  cout << slope << endl;

And now we have a complete solution!


Christopher W Brown
Last modified: Wed Aug 11 16:09:29 EDT 2004