Question

Problem #1 Write a user-defined function called mysecant. (Not an an please!) The purpose of the function is: Given two points, compute the line between them and so where the corresponding y3-0. If there is no solution, Hint: The governing equations are: (y3-y1)-(y2-y1) / (x2-x1) * (x3-x1) if y3-0, then solve for x3, and so, x3-x1-y1 * (x2-x1) / (y2-y Inputs: x1, y1, x2, y2 Output: x3

Answer 1

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/*HomeworkLib Quetion */
#include <stdio.h>
float mysecant(float x1,float y1, float x2, float y2)
{
float x3;
x3=x1-y1*(x2-x1)/(y2-y1); //if y3=0 than
return x3;
}

int main()
{
float x1,y1,x2,y2,x3,y3=0;
printf("Enter Point 1(x1,y1):"); //input of point 1
scanf("%f %f",&x1,&y1);
printf(" Enter Point 2(x2,y2):"); //input for point 2
scanf("%f %f",&x2,&y2);
  
x3=mysecant(x1,y1,x2,y2); //call function mysecant
  
printf(" (x3,y3)=(%.2f,0)",x3); //print result
  
return 0;
}

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