Editing Perpendicular (section)

== Graph of functions ==
[[File:Slopes and orthogonality.svg|thumb|upright=.95|Two perpendicular lines have slopes {{math|1=''m''<sub>1</sub> = Δ''y''<sub>1</sub>/Δ''x''<sub>1</sub>}} and  {{math|1=''m''<sub>2</sub> = Δ''y''<sub>2</sub>/Δ''x''<sub>2</sub>}} satisfying the relationship  {{math|1=''m''<sub>1</sub>''m''<sub>2</sub> = &minus;1}}.]]
In the two-dimensional plane, right angles can be formed by two intersected lines if the [[Product (mathematics)|product]] of their [[slopes]] equals  −1. Thus for two [[linear function]]s <math>y_1(x) = m_1 x + b_1</math> and <math>y_2(x) = m_2 x + b_2</math>, the graphs of the functions will be perpendicular if <math>m_1 m_2 = -1.</math> 

The [[dot product]] of [[Euclidean vector|vector]]s can be also used to obtain the same result: First, [[Translation of axes|shift coordinates]] so that the origin is situated where the lines cross. Then define two displacements along each line, <math>\vec r_j</math>, for <math>(j=1,2).</math> Now, use the fact that the inner product vanishes for perpendicular vectors:
:<math>\vec r_1=x_1\hat x + y_1\hat y =x_1\hat x + m_1x_1\hat y</math>
:<math>\vec r_2=x_2\hat x + y_2\hat y = x_2\hat x + m_2x_2\hat y</math>
:<math>\vec r_1 \cdot \vec r_2 = \left(1+m_1m_2\right)x_1x_2 =0</math>
:<math>\therefore m_1m_2=-1</math> (unless  <math>x_1</math> or <math>x_2</math> vanishes.)

Both proofs are valid for horizontal and vertical lines to the extent that we can let one slope be <math>\varepsilon</math>, and take the limit that <math>\varepsilon\rightarrow 0.</math>  If one slope goes to zero, the other goes to infinity.