ABSTRACT
So far, we have seen that for points on a plane we can express a straight line as a series of values (x, y) such that
ax + by + c = 0
We can also express it in the form that
y = mx + n
where m is the slope or gradient of the line and n is a constant. These two expressions represent the same line if
m = –a/b and n = –c/b
We also saw in Chapter 6 and can see from Figure 9.1 that if the slope of the line AB is θ (measured from the horizontal axis anticlockwise), then
y = x tan θ + c
Here, c is the value of y when x is zero. We can also express this as dy/dx = tan θ = m, which is constant for a straight line since
∫ tan θ dx = x tan θ + c
The differential of a constant is zero; hence, for a line, d2y/dx2 = 0. Also note that if we use clockwise bearings from the vertical rather than anti-
clockwise angles from the horizontal, the bearing of line AB = (90 – θ) and
y = x cot (Bearing AB) + c
If the line CD is perpendicular to AB, its slope will be (90 + θ). Hence, for the line CD
y = m′ x + c′
where
m′ = –cot (θ) = –tan (Bearing AB)
We have also seen that when two lines AB and CD are at right angles, their slopes are such that
m m′ = tan θ * (–cot θ) = –1
If the product of the slopes of two lines = –1 they are said to be orthogonal. Thus, in Figure 9.1, the lines AB and CD are orthogonal. In Chapter 7, we also used the term orthogonal for a matrix where A A-1 = I.
