Gradients, parallel and perpendicular lines

For two distinct points, \(m=\dfrac{y_2-y_1}{x_2-x_1}\) when \(x_2\ne x_1\). A vertical line is \(x=k\) and has undefined gradient.

The general form \(ax+by+d=0\) includes vertical lines. For \(b\ne0\), rearrange to \(y=-\dfrac{a}{b}x-\dfrac{d}{b}\).

Example

Through \((2,5)\) perpendicular to \(y=2x-1\): \(m=-\dfrac12\), so \(y-5=-\dfrac12(x-2)\), or \(x+2y-12=0\).

For points \(A(x_1,y_1)\) and \(B(x_2,y_2)\), \(AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) and the midpoint is \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\). Distance is nonnegative; gradient is signed. Average each coordinate separately.

For a quadrilateral, equal opposite-side gradients establish parallel sides. To establish a rectangle, show a parallelogram and a right angle; one pair of perpendicular sides alone is insufficient.

For \(A(-3,-2)\) and \(B(1,1)\), the horizontal change is \(4\) and the vertical change is \(3\). Hence the gradient is \(\dfrac34\) and the distance is \(\sqrt{4^2+3^2}=5\).