Line equations
Equations chart
Adjust the blue slider to modify \(m\) , the slope, and the red slider to change \(b\), the \(y\)-intercept.
Line characteristics
Move the points \(A\) and \(B\) to change the segment coordinates.
You can find the distance between points \(A \left(x_{1},y_{1}\right)\) and \(B \left(x_{2},y_{2}\right)\) by using the next equation:
$$D = \sqrt{ \left(x_{2} - x_{1}\right)^{2} + \left(y_{2} - y_{1}\right)^{2} }$$
Equations
| name | equation |
|---|---|
| slope intercept form | $$y = mx + b$$ |
| point-slope form | $$y - y_{1} = m(x - x_{1})$$ |
| two points form | $$y - y_{1} = \frac{\left(y_{2} - y_{1}\right)}{\left(x_{2} - x_{1}\right)} \left(x - x_{1}\right)$$ |
| general form | $$Ax + By + C = 0$$ |
| intercept form | $$\frac{x}{a} + \frac{y}{b} = 1$$ |
| determinant form (1) | $$\left( x_{2} - x_{1} \right)\left( y - y_{1} \right) - \left(y_{2} - y_{1}\right)\left(x - x_{1}\right) = 0$$ $$\begin{vmatrix} x - x_{1} & y - y_{1} \\ x_{2} - x_{1} & y_{2} - y_{1} \end{vmatrix} = 0$$ |
| determinant form (2) | $$\left( y_{1} - y_{2} \right)x + \left( x_{2} - x_{1} \right)y + \left(x_{1}y_{2} - x_{2}y_{1}\right) = 0$$ $$\begin{vmatrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{vmatrix} = 0$$ |
| parametric form | $$x = x_{0} + at$$ $$y = y_{0} + bt$$ |
| distance | $$D = \sqrt{ \left(x_{2} - x_{1}\right)^{2} + \left(y_{2} - y_{1}\right)^{2} }$$ |
| slope | $$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$$ |
| angle between lines | $$\theta = \arctan \left( \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} \right)$$ |
| midpoint | $$\begin{split}P_{m} &= \left(x_{m},y_{m}\right) \\ &= \left(\frac{x_{1} + x_{2}}{2},\frac{y_{1} + y_{2}}{2}\right)\end{split}$$ |
| distance between a point and a line | $$d \left( \left( x_{0},y_{0} \right), Ax + By + C \right) = \frac{\left\vert Ax_{0} + By_{0} + C \right\vert}{\sqrt{A^{2} + B^{2}}}$$ |
Calculations
Intercept form
To obtain the intercept form from the slope intercept form substitute \(m = -\frac{b}{a}\)
$$\begin{split} mx + b &= y \\ -\frac{b}{a} x + b &= y \\ \frac{b}{a}x + y &= b \\ \frac{x}{a} + \frac{y}{b} &= 1 \end{split}$$
General form to point-slope
The relationship between the general equation and the point-slope equation is:
$$\begin{split} Ax + By + C &= 0 \\ By &= Ax - C \\ \Rightarrow y &= -\frac{A}{B}x + \left( -\frac{C}{B} \right) \end{split}$$
So,
$$m = -\frac{A}{B}$$
$$b = -\frac{C}{B}$$
Point-slope form to general
To get the general form from the slope intercept form
$$\begin{split}y &= mx + b \\ 0 &= mx - y + b \end{split}$$
General form to parametric form
Find a point \(P_{0}(x_{0},y_{0})\) on the line, choose convenient values for \(x\) or \(y\) and solve the equation. For example, if \(B \neq 0\), set \(x = 0\) and solve for \(y\), or if \(A \neq 0\), set \(y = 0\) and solve for \(x\).
A direction vector \( \vec{v} = (a,b) \) of the line can be obtained by swapping the coefficients \(A\) and \(B\) and changing the sign of one of them. A possible direction vector is \((b,-a)\) because it follows the directional relationship of the line.
Construct the parametric equations using the form:
$$x = x_{0} + Bt$$
$$y = y_{0} - At$$
where \((x_{0},y_{0})\) is the chosen point and \((a,b)\) is the direction vector.