This equation represents a general form for the equation of a straight line, given it's gradient (\( \htmlClass{sdt-0000000138}{m} \)), and where it crosses the \( \htmlClass{sdt-0000000017}{y} \) axis.
| \( x \) | This is a symbol for any generic variable. It can hold any value, whether that be an integer or a real number, or a complex number, or a matrix etc. |
| \( y \) | This is a secondary symbol for any generic variable. It can hold any value, whether that be an integer or a real number, or a complex number, or a matrix etc. |
| \( C \) | This is the symbol for the 'y intercept' of a straight line. In the case of a line on the \(xy\)-plane, \(\htmlClass{sdt-0000000096}{f}(\htmlClass{sdt-0000000003}{x})\), it is the value of \(\htmlClass{sdt-0000000096}{f}(0)\) |
| \( m \) | This is the symbol for the gradient of a straight line. It is a ratio between a change in the input, and a change in the output. |
Coming soon...
Finding the equation for the straight line in the following figure:
We can see from the image that the \(y\) intercept is equal to \(1\).
We can then take the starting point (as presented in the figure) \((-3, -5)\) and the ending point \((2, 5)\) to calculate the gradient.
We can now use the fact that:
\[ \htmlClass{sdt-0000000138}{m} = \frac{\Delta \htmlClass{sdt-0000000017}{y}}{\Delta \htmlClass{sdt-0000000003}{x}} \]
and say that:
\[ \htmlClass{sdt-0000000138}{m} = \frac{5--5}{2--3} = \frac{10}{5} = 2 \]
We can now plug these values into our equation to get the final form:
\[ \htmlClass{sdt-0000000017}{y} = 2 \htmlClass{sdt-0000000003}{x} + 1 \]