Perpendicular



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. Note that by definition, a line is infinitely long, and strictly speaking AB and CD in this example represent line segments of two infinitely long lines. Hence the line segment AB does not have to intersect line segment CD to be considered perpendicular lines, because if the line segments are extended out to infinity, they would still form congruent adjacent angles.

If a line is bending to another as in Figure 1, all of the angles created by their intersection are called right angles (right angles measure ½&pi; radians, or 90°). Conversely, any lines that meet to form right angles are perpendicular.

In a coordinate plane, perpendicular lines have opposite reciprocal slopes. A horizontal line has slope equal to zero while the slope of a vertical line is described as undefined or sometimes ±infinity. Two lines that are perpendicular would be denoted as.

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,.

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). 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.
 * 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.

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'.

Algebra
In algebra, for any linear equation y=mx + b, the perpendiculars will all have a slope of (-1/m), the opposite reciprocal of the original slope. It is helpful to memorize the slogan "to find the slope of the perpendicular line, flip the fraction and change the sign." Recall that any whole number a is itself over one, and can be written as (a/1)

To find the perpendicular of a given line which also passes through a particular point (x, y), solve the equation y = (-1/m)x + b, substituting in the known values of m, x, and y to solve for b.

Calculus
First find the derivative of the function. This will be the slope (m) of any curve at a particular point (x, y). Then, as above, solve the equation y = (-1/m)x + b, substituting in the known values of m, x, and y to solve for b.