Any straight line can be written in the standard form: $$y=mx+c$$
The gradient, $m$ is given by the formula: $$m=\frac{y_2-y_1}{x_2-x_1}$$
The y-intercept, $c$ can be found by substituting the gradient and any of the points (in this case we use $(x_1,y_1)$) into the standard form of the line equation:
$$ \begin{equation}\begin{aligned} y_1&=mx_1+c\\ \therefore c&=y_1-mx_1\\ \end{aligned}\end{equation} $$The following code snippet shows how we can replicate this in C:
#include <stdio.h>
int main() {
// storing the points
double x1 = -2.5, x2 = 0, y1 = 0, y2 = 5;
printf("Using the points (%g,%g) and (%g,%g)\n", x1, y1, x2, y2);
// calculating gradient
double gradient = (y2 - y1) / (x2 - x1);
printf("The gradient of the line is %g\n", gradient);
// calculating the y-intercept
double y_intercept = y1 - gradient * x1;
printf("The y intercept is %g\n", y_intercept);
// displaying the final equation of the straight line
printf("Therefore the equation of the line is:\ny=%gx+%g", gradient,
y_intercept);
return 0;
}
finding-the-equation-of-a-line-given-two-points.c Copy