an equation of the first degree satisfied by x and y. No point not on the line satisfies the same equation; for the line from C to any point off the line would make with CR some angle β different from α, and the point in question would satisfy an equation y − k = tan β(x − h), which is inconsistent with the above equation.
The equation of the line may also be written y = mx + b, where m = tan α, and b = k − h tan α. Here b is the value obtained for y from the equation when 0 is put for x, i.e. it is the numerical measure, with proper sign, of OB, the intercept made by the line on the axis of y, measured from the origin. For different straight lines, m and b may have any constant values we like.
Now the general equation of the first degree Ax + By + C = 0 may be written y = −(A/B)x − C/B, unless B = 0, in which case it represents a line parallel to the axis of y; and −A/B, −C/B are values which can be given to m and b, so that every equation of the first degree represents a straight line. It is important to notice that the general equation, which in appearance contains three constants A, B, C, in effect depends on two only, the ratios of two of them to the third. In virtue of this last remark, we see that two distinct conditions suffice to determine a straight line. For instance, it is easy from the above to see that
| x | + | y | = 1 |
| a | b |
is the equation of a straight line determined by the two conditions that it makes intercepts OA, OB on the two axes, of which a and b are the numerical measures with proper signs: note that in fig. 50 a is negative. Again,
| y − y1 = | y2 − y1 | (x − x1), |
| x2 − x1 |
i.e.
(y1 − y2) x − (x1 − x2) y + x1y2 − x2y1 = 0,
represents the line determined by the data that it passes through two given points (x1, y1) and (x2, y2). To prove this find m in the equation y − y1 = m(x − x1) of a line through (x1, y1), from the condition that (x2, y2) lies on the line.
In this paragraph the coordinates have been assumed rectangular. Had they been oblique, the doctrine of similar triangles would have given the same results, except that in the forms of equation y − k = m(x − h), y = mx + b, we should not have had m = tan α.