My Formula for slope-intercept Form of a Line

Goal: given two points (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2), find the straight line in R2\R^2 that connects these two points in the slope-intercept form in terms of the x1x_1, y1y_1, x2x_2 and y2y_2.

y=mx+cy=mx+c

m=ΔyΔx=y1y2x1x2m=\frac{\Delta y}{\Delta x}=\frac{y_1-y_2}{x_1-x_2}

y=mx+cy1=mx1+cc=y1mx1=y1(y1y2x1x2)x1=x1y1x2y1x1x2x1y1x1y2x1x2=x1y1x2y1x1y1+x1y2x1x2=x1y2x2y1x1x2c=1x1x2x1y1x2y2\begin{aligned} y&=mx+c\\ y_1&=mx_1+c\\ c&=y_1-mx_1\\ &=y_1-\left(\frac{y_1-y_2}{x_1-x_2}\right)x_1\\ &=\frac{x_1y_1-x_2y_1}{x_1-x_2}-\frac{x_1y_1-x_1y_2}{x_1-x_2}\\ &=\frac{x_1y_1-x_2y_1-x_1y_1+x_1y_2}{x_1-x_2}\\ &=\frac{x_1y_2-x_2y_1}{x_1-x_2}\\ c&=\frac{1}{x_1-x_2}\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix} \end{aligned}

y=(y1y2x1x2)x+1x1x2x1y1x2y2\therefore\boxed{y=\left(\frac{y_1-y_2}{x_1-x_2}\right)x+\frac{1}{x_1-x_2}\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}}

#Algebra

Note by Gandoff Tan
1 year, 5 months ago

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