The equation of a straight line takes the form y=mx+c, where m is the gradient (slope) of the line and c the y-intercept (The value of y where the line crosses the y-axis). The values of m and c can be calculated if two points on the line are known. If these points are denoted by (x_{1},y_{1}) and (x_{2},y_{2}), then y_{1}=mx_{1}+c and y_{2}=mx_{2}+c. From this, it follows that y_{2}-y_{1}=m(x_{2}-x_{1}). The value of m can now be found. To find c, insert the known values into the equation for y_{1} or y_{2} and rearrange.

A quadratic equation is one of the form y=ax²+bx+c, where a, b and c denote constants. The resulting curve is a parabola. Equations of the form y=f(x) in which f(x) consists only of powers of x and a pure number are termed polynomials. Polynomial equations in which the highest power of x is 3, 4 or 5 are cubic, quartic and quintic equations respectively.

Consider a circle of radius r and centre (0,0). If the radius is the hypotenuse of a right-angled triangle (The side not touching the right angle), and (x,y) denotes a point on the circumference, Pythagoras' Rule states that x²+y²=r². Now move the centre to (a,b). The x coordinate, to take an example, of the point (x,y) is now *a* units greater than the length of the horizontal side of the triangle. To return to the dimensions of the triangle, therefore, the equation of the circle must be written as (x-a)²+(y-b)²=r².

Imagine that the equation is written in the form x²+ax+y²+by=c. This can be rewritten as (x+0.5a)²-0.25a²+(y+0.5b)²-0.25b²=c, then as (x+0.5a)²+(y+0.5b)²=0.25a²+0.25b²+c, where 0.25a²+0.25b²+c=r². This technique is called *completing the square*.

The equation of an ellipse is of the form x²/a²+y²/b²=1. The centre of this ellipse is at (0,0), so the foci can be denoted by (-α,0) and (α,0). The lines d_{1} and d_{2} connect the foci to a point on the ellipse. The sum of the lengths of these lines is constant, so let us consider the points where the ellipse crosses the positive axes. At (a,0), d_{1}=a+α and d_{2}=a-α. Therefore, d_{1}+d_{2}=2a. At (0,b), d_{1}=d_{2}=√(α²+b²). Therefore, d_{1}+d_{2}=2√(α²+b²). Combining the equations gives 2a=2√(α²+b²), and rearranging gives α=√(a²-b²). Another equation is α=a∊, where ∊ denotes the eccentricity.

The equation of a hyperbola is of the form x²/a²-y²/b²=1. Rearranging this gives y=(b/a)√(x²-a²). The hyperbola gets closer to the asymptotes (The lines it does not cross) as x→∞ or -∞. There, √(x²-a²)≈√(x²), which in turn equals x or -x. As our hyperbola is centred at (0,0), the gradients of the asymptotes are therefore b/a and -b/a. A hyperbola is rectangular if a=b.

To find the intersection(s) of two lines, one can start by writing both equations in terms of the same variable, for example x. This gives equations of the form x=f_{1}(y) and x=f_{2}(y). If two quantities both equal x (or the same function of x), they must also be equal to each other. Therefore, f_{1}(y)=f_{2}(y). Now write this equation in terms of y. This technique only works where both the x and y values are the same for both lines, so one now has the y coordinate(s) of the intersection(s). Finally, substitute for y in either of the original equations to find the x coordinate(s).