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Using Properties of Perpendicular Lines


Perpendicular Lines

A pair of lines that intersect at a right angle are said to be perpendicular.


Name all pairs of perpendicular lines in the diagram.

Show Solution

We can begin by explicitly noting the given information in the diagram. First, we know that ab a \parallel b — they aren't perpendicular. Also given is that ac,a \perp c, which is our first pair of perpendicular lines. This information about a,a, b,b, and cc can help us to decide if bb and cc are perpendicular. As ab,a \parallel b, the angles at the intersection of bb and cc are corresponding to the ones at the intersection of aa and c.c.

Thus, bb and cc are also a pair of perpendicular lines. Lastly, is dd perpendicular to any of the other lines? Looking at the diagram, we can see that it's not. However, this can also be argued mathematically. If dc d \perp c was true, dd would also be parallel to aa and b.b. Similarly, if it was perpendicular to aa and b,b, it would be parallel to c.c. As dd intersects all other lines, it's not parallel to any of them. Thus, it can't possibly be perpendicular to any line either.



There are two pairs of perpendicular lines in the diagram: aca \perp c and bc.b \perp c.


Perpendicular Postulate

The perpendicular postulate states that for a point PP not on the line l,l, there exists only one line perpendicular to ll that passes through P.P.

This postulate is one of the very basic truths used when proving and finding further characteristics of perpendicular lines.


Drawing a Perpendicular Line


Given a point AA on the line m,m, it is possible to use a compass and straightedge to create a line that is perpendicular to mm and that passes through A.A. This will be the unique line described by the Perpendicular Postulate.

First, place the sharp end of the compass at point A,A, and draw a circle.
Draw circle

The circle intersects mm at two distinct points. These points will be named XX and Y,Y, for later referencing.

Place the sharp end of the compass at XX and draw an arc above point A.A.
Draw arc

Next, using the same compass setting, the sharp end is placed at Y,Y, and another arc is drawn above AA such that it intersects the first arc.
Draw arc

By using a straightedge to draw a line through AA and the intersection of the arcs, the desired perpendicular line is constructed.
Draw Line


Shortest Distance from a Point to a Line

The shortest distance from a point PP to a line ll is always along the line perpendicular to ll through P.P.


Find the shortest distance from the point PP to the line l.l.

Show Solution

The shortest route from PP to ll is along the line through PP perpendicular to l.l. To find the desired distance, we have to first find where this perpendicular line intersects l.l. Perpendicular lines, expressed as linear functions, satisfy the following condition. m1m2=-1 m_1 \cdot m_2 = \text{-} 1 Thus, the slope of shortest route from PP to ll is the negative reciprocal of the slope of l.l. Let's find the slope of l.l. Increasing xx by 11 increases the value of yy by 2.2. Therefore, the slope of ll is 2,2, giving us the desired slope -12=-0.5. \text{-} \dfrac 1 2 = \text{-} 0.5. We can now draw a line through PP with the slope -0.5.\text{-} 0.5.

These two lines intersect at (1,2),(1,2), so the shortest distance from PP to ll is the distance from P,P, at (3,1),(3,1), to (1,2).(1,2). We'll calculate this using the distance formula.

d=(x2x1)2+(y2y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 }
d=(31)2+(12)2d = \sqrt{\left({\color{#0000FF}{3}}-{\color{#009600}{1}}\right)^2 + \left({\color{#0000FF}{1}}-{\color{#009600}{2}}\right)^2}
d=22+(-1)2d = \sqrt{2^2 + (\text{-} 1)^2}
d=4+1d = \sqrt{4 + 1}
d=5d = \sqrt{5}

The shortest distance between PP and ll is 5\sqrt{5} units.


Perpendicular Bisector

The perpendicular bisector of a segment AB\overline{AB} is the line that is perpendicular to AB\overline{AB} and intersects AB\overline{AB} at its midpoint.


Drawing a Perpendicular Bisector


Given the segment AB,\overline{AB}, a compass and straightedge can be used to draw its perpendicular bisector.

Place the compass' sharp end at one of the segment's endpoints. Draw an arc with a radius larger than half the distance between AA and B.B.

Keeping the compass length the same, draw a corresponding arc on the opposite side. The two arcs should now intersect at two distinct points.

The line that contains these two intersections can now be drawn using a straightedge.

This line is perpendicular to AB,\overline{AB}, and their intersection is at the midpoint of AB.\overline{AB}. Thus, it is the desired perpendicular bisector.

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