Gradients and lines in shapes

Gradient = change in y ÷ change in x. Two sides with the same gradient are parallel. In a parallelogram ABCD, the side CD has the gradient of AB and passes through C, which gives its equation.

1. A(1, 1), B(3, 5), C(6, 4) are three corners of parallelogram ABCD
2. gradient of AB = (5 − 1) ÷ (3 − 1) = 4 ÷ 2 = 2
3. CD: y = 2x + c through C(6, 4): 4 = 12 + c, so c = −8
4. y = 2x − 8

Join the midpoints M of AB and N of AC in any triangle ABC: MN always has the gradient of BC, and BC is twice as long. Work out M and N, then compare the steps.

A corner lies on both of its sides, so its coordinates satisfy both equations. Set the right-hand sides equal and solve.

1. y = 2x + 4 and y = −x + 7
2. 2x + 4 = −x + 7, so 3x = 3 and x = 1
3. y = 2 × 1 + 4 = 6, so they meet at (1, 6)
4. with the x-axis: y = 0 gives x = −2 and x = 7, a base of 9
5. area of the triangle = ½ × 9 × 6 = 27 square units

If one side has gradient m, a side at right angles to it has gradient −1/m: turn the gradient over and change its sign. So 2 and −1/2 are perpendicular, and so are −2/5 and 5/2. This proves a right angle in a shape: work out the two gradients and check that their product is −1.

The height of a triangle above a side AB is the line from the opposite corner C at right angles to AB. Its gradient is −1 ÷ the gradient of AB, and it passes through C. Solve it with the equation of AB to find the foot F; then CF, by Pythagoras, is the height.

The diagonals of a rhombus cross at right angles, at the midpoint of each one. So BD is the line through the midpoint of AC with the perpendicular gradient.

In a square, the next side is the same length and at right angles. If the step from A to B is p across and q up, a quarter turn anticlockwise makes it −q across and p up. The gradients q/p and −p/q multiply to −1, and the lengths are equal.

1. A(2, 1) to B(6, 2): 4 across, 1 up
2. quarter turn anticlockwise: 1 left, 4 up
3. C = (6 − 1, 2 + 4) = (5, 6)
4. D = (2 − 1, 1 + 4) = (1, 5)