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In geometry, two lines or planes (or a line and a plane), are considered 'perpendicular' (or 'orthogonal') to each other if they form congruent adjacent angles. The term may be used as a noun or adjective. Thus, referring to Figure 1, the line AB is the 'perpendicular' to CD through the point B.
If a line is perpendicular to another as in Figure 1, all of the angles created by their intersection are called ''right angles'' (right angles measure ½π radians, or 90°). Conversely, any lines that meet to form right angles are perpendicular. The line AB does not have to end at B to be considered perpendicular.
In a co-ordinate plane, perpendicular lines have opposite reciprocal slopes. Horizontal and vertical lines have zero and positive/negative infinity.

Contents

Numerical criteria

In terms of slopes

Construction of the perpendicular

In relationship to parallel lines

See also

External links

Numerical criteria

In terms of slopes

In a Cartesian coordinate system, two straight lines $L$ and $M$ may be described by equations.
:$L\; :\; y\; =\; ax\; +\; b,$
:$M\; :\; y\; =\; cx\; +\; d,$
as long as neither is vertical. Then $a$ and $c$ are the slopes of the two lines. The lines $L$ and $M$ are perpendicular if and only if the product of their slopes is -1, or if $ac=-1$.
The perpendiculars to vertical lines are always horizontal lines, and the perpendiculars to horizontal lines are always vertical lines. All horizontal lines are perpendicular to all vertical lines; that is, for any horizontal line $P\; :\; x\; =\; J$ and horizontal line $Q\; :\; y\; =\; K$, where $J$ and $K$ are constants, $P\; perp\; Q$.

Construction of the perpendicular

To construct the perpendicular to the line AB through the point P using compass and straightedge, proceed as follows (see Figure 2).

★ Step 1 (red): construct a circle with center at P to create points A' and B' on the line AB, which are equidistant from P.

★ Step 2 (green): construct circles centered at A' and B', both passing through P. Let Q be the other point of intersection of these two circles.

★ Step 3 (blue): connect P and Q to construct the desired perpendicular PQ.
To prove that the PQ is perpendicular to AB, use the SSS congruence theorem for triangles QPA' and QPB' to conclude that angles OPA' and OPB' are equal. Then use the SAS congruence theorem for triangles OPA' and OPB' to conclude that angles POA and POB are equal.

In relationship to parallel lines

As shown in Figure 3, if two lines (''a'' and ''b'') are both perpendicular to a third line (''c''), all of the angles formed on the third line are right angles. Therefore, in Euclidean geometry, any two lines that are both perpendicular to a third line are parallel to each other, because of the parallel postulate. Conversely, if one line is perpendicular to a second line, it is also perpendicular to any line parallel to that second line.
In Figure 3, all of the orange-shaded angles are congruent to each other and all of the green-shaded angles are congruent to each other, because vertical angles are congruent and alternate interior angles formed by a transversal cutting parallel lines are congruent. Therefore, if lines ''a'' and ''b'' are parallel, any of the following conclusions leads to all of the others:

★ One of the angles in the diagram is a right angle.

★ One of the orange-shaded angles is congruent to one of the green-shaded angles.

★ Line 'c' is perpendicular to line 'a'.

★ Line 'c' is perpendicular to line 'b'.

See also

★ Orthogonality

★ Perpendicular component (of a vector)

★ Surface normal

External links

★ Definition: perpendicular With interactive animation

★ How to draw a perpendicular bisector of a line with compass and straight edge Animated demonstration

★ [http://www.mathopenref.com/constperpendray.html How to draw a perpendicular at the endp[oint of a ray with compass and straight edge] Animated demonstration