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Lesson Explainer: Geometric Constructions: Perpendicular Lines Mathematics

In this explainer, we will learn how to construct, using a ruler and a pair of compasses, the perpendicular to a given line from or at a given point and the perpendicular bisector of a line segment.

We first note that for any line 𝐴𝐵, there is a unique line that is perpendicular to 𝐴𝐵 that passes through a given point 𝐶𝐴𝐵. We can construct this line with a compass and a straight edge by using our knowledge of circles and triangles.

We start by setting the point of our compass at 𝐶, and then we can use any size for the radius. Showing only the arcs that intersect gives us the following.

We can call the points of intersection between this circle and 𝐴𝐵 𝐴 and 𝐵 respectively. We note that 𝐴𝐶=𝐵𝐶 since these are both radii of the same circle, so we can say that 𝐶 is equidistant from 𝐴 and 𝐵. If we then set the radius of the compass to be larger than 𝐴𝐶, we can trace circles (or just their arcs) centered at 𝐵 and 𝐴 such that they intersect twice. We will call the point of intersection above 𝐴𝐵 𝐷, as shown.

We then note that 𝐵𝐷=𝐴𝐷, as they are radii of congruent circles. We can add these lines along with 𝐶𝐷 onto the diagram.

We now note that triangles 𝐵𝐶𝐷 and 𝐴𝐶𝐷 have congruent sides, so by the SSS criterion, they are congruent triangles. In particular, this means that 𝐴𝐶𝐷𝐵𝐶𝐷. For two congruent angles to make up a straight angle, their measures must sum to 180, so they are both right angles.

Hence, we have constructed 𝐶𝐷, which is perpendicular to 𝐴𝐵 passing through 𝐶.

We can follow this process to find the perpendicular line passing through any given point on a line.

How To: Constructing the Perpendicular Line through a Given Point on the Line

To find the perpendicular line to 𝐴𝐵 passing through 𝐶𝐴𝐵, we use the following steps:

  1. We start by setting the point of our compass at 𝐶 and then trace a circle intersecting 𝐴𝐵 twice. We can label these points 𝐴 and 𝐵.
  2. We then set the radius of the compass to be larger than 𝐴𝐶 and we can trace arcs of circles centered at 𝐵 and 𝐴 such that they intersect. We will call the point of intersection above 𝐴𝐵 𝐷, as shown.
  3. We have that 𝐶𝐷𝐴𝐵.

This allows us to find the perpendicular line passing through any given point on a line. However, there is a special case of this that splits 𝐴𝐵 into two equal parts called the perpendicular bisector. We define this formally as follows.

Definition: Bisectors and Perpendicular Bisectors

To bisect a line segment means to separate it into two equal parts.

In particular, the perpendicular bisector of a line segment is the line that separates the line segment into two equal parts and meets the line segment at right angles.

We can follow a similar process to construct the perpendicular bisector of any line segment 𝐴𝐵. We start by setting the radius of the compass to be greater than 𝐴𝐵 and then sketching arcs of circles of this radius centered at 𝐴 and 𝐵. These will intersect at two points that we will label 𝐶 and 𝐷, as shown.

Connecting 𝐶 and 𝐷 with a straight line and adding in the radii gives us the following.

We note that 𝐶𝐵𝐷 and 𝐶𝐴𝐷 have three congruent sides, so by the SSS criterion, they are congruent. Therefore, their corresponding angles are congruent. So, 𝑚𝐵𝐶𝐷=𝑚𝐴𝐶𝐷. We can add this onto the diagram along with labeling the point of intersection between 𝐴𝐵 and 𝐶𝐷 𝐸.

We then see that 𝐴𝐸𝐶 and 𝐵𝐸𝐶 share two congruent sides and the included angles are also congruent. Thus, 𝐴𝐸𝐶𝐵𝐸𝐶 by the SAS criterion.

Hence, their corresponding sides are all congruent, and so 𝐴𝐸=𝐵𝐸. We also note that 𝑚𝐴𝐸𝐶=𝑚𝐵𝐸𝐶 and they sum to make a straight angle, so they must be right angles. It is also worth noting that this proves the diagonals of a rhombus meet at right angles.

This means that 𝐶𝐷 is the perpendicular bisector of 𝐴𝐵. We can follow this process in general to construct the perpendicular bisector of any line segment.

How To: Constructing the Perpendicular Bisector of a Line Segment

To find the perpendicular bisector to 𝐴𝐵, we use the following steps:

  1. Set the radius of the compass to be greater than 𝐴𝐵.
  2. Trace arcs of two circles of this radius centered at 𝐴 and 𝐵. These will intersect at two points that we will name 𝐶 and 𝐷.
  3. 𝐶𝐷 is then the perpendicular bisector of 𝐴𝐵.

Let’s now see an example of determining the relationship a geometric construction illustrates.

Example 1: Constructing the Perpendicular Bisector of a Line Segment

Which of the following does the figure illustrate?

What does the following figure illustrate

  1. A perpendicular to a straight line originating from it
  2. A bisector of an angle
  3. A perpendicular bisector of a line segment
  4. A straight line parallel to another line
  5. A perpendicular from a point lying outside a straight line

Answer

We can answer this question by recalling that the perpendicular bisector of a line segment 𝐴𝐵 is constructed by the line segment between the intersection of circles centered at 𝐴 and 𝐵. This is exactly what is pictured, so the answer is option C: a perpendicular bisector of a line segment.

In our next example, we will see another question on determining the relationship a given geometric construction illustrates.

Example 2: Constructing the Perpendicular to a Line from a Point on the Line

What does the following figure illustrate?

  1. A bisector of an angle
  2. A perpendicular to a straight line originating from it
  3. A perpendicular from a point lying outside a straight line
  4. A straight line parallel to another line
  5. A bisector of a line segment

Answer

We can answer this question by recalling that the perpendicular bisector of a line segment 𝐴𝐵 through a point on the line 𝐶 is constructed by first tracing a circle at 𝐶 to find two points on 𝐴𝐵 equidistant from 𝐶—say 𝐴 and 𝐵. Then, we trace circles at 𝐴 and 𝐵 to find a second point equidistant from 𝐴 and 𝐵—say 𝐷. We then recall that the line segment 𝐶𝐷 is perpendicular to 𝐴𝐵.

This is exactly what is pictured, so the answer is option B: a perpendicular to a straight line originating from it.

Before we move on to the example, there is one more construction we need to consider. We have seen how to find the perpendicular line to 𝐴𝐵 through a point 𝐶 on the line, but what if 𝐶 does not lie on the line?

We start by tracing a circle centered at 𝐶 that intersects 𝐴𝐵 twice; we label these points 𝐴 and 𝐵.

We note that 𝐶 is equidistant from 𝐴 and 𝐵, since 𝐶𝐴 and 𝐶𝐵 are radii of the circle centered at 𝐶. We can find another point equidistant from 𝐴 and 𝐵 by tracing circles of the same radius centered at each point that intersect at a point 𝐷, as shown.

Since we kept the radii of the circles the same, 𝐵𝐷=𝐴𝐷=𝐵𝐶=𝐴𝐶, adding these congruencies and the line segment 𝐶𝐷 to the diagram gives us the following.

This is the same construction we had previously, where 𝐶𝐷 and 𝐴𝐵 are diagonals of a rhombus. Thus, we can conclude they meet at right angles.

We can follow this process in general to construct the line perpendicular to any line through a point not on the line.

How To: Constructing the Perpendicular of a Line through a Point Not on the Line

To find the perpendicular to 𝐴𝐵 passing through 𝐶𝐴𝐵, we use the following steps:

  1. Trace a circle centered at 𝐶 that intersects 𝐴𝐵 twice; label these points 𝐴 and 𝐵.
  2. Trace circles centered at 𝐴 and 𝐵 that intersect; label the point of intersection 𝐷.
  3. 𝐶𝐷 is then perpendicular to 𝐴𝐵.

Let’s now see an example of using this construction to identify which construction represents drawing a perpendicular from a point lying outside a straight line.

Example 3: Identifying the Construction of the Perpendicular to a Line from a Point Lying outside the Line

Which of the following constructions represents drawing a perpendicular from a point lying outside a straight line?

Answer

We recall that to find a perpendicular from a point 𝐶 lying outside the straight line 𝐴𝐵, we first trace a circle at 𝐶 that intersects 𝐴𝐵 at two distinct points—say 𝐷 and 𝐹. If we then trace circles centered at 𝐷 and 𝐹 that intersect at a point 𝐸, then 𝐶𝐸𝐴𝐵.

We can see that this option E.

It is worth noting that the other options give different geometric constructions. Option A is a construction of a perpendicular bisector of a line segment. Option B finds the perpendicular line to a straight line passing through a point on the line. Option C is a bisector; however, it is not perpendicular. And option D is beyond the scope of this explainer, but it is a construction of an angle bisector.

So, the answer is option E.

Before we move on to our next example, we can show a few useful properties of the perpendicular bisector of a line segment.

Consider the following line segment 𝐴𝐵 with perpendicular bisector 𝐶𝐷, as shown.

We can first show that every point on 𝐶𝐷 is equidistant from 𝐴 and 𝐵. To do this, let’s consider any point 𝐸𝐶𝐷. We can construct triangle 𝐴𝐵𝐸 as shown.

We note that triangles 𝐴𝐶𝐸 and 𝐵𝐶𝐸 share two congruent sides and the included angles are both right angles. So, they are congruent. Hence, 𝐵𝐸=𝐴𝐸.

We now note that since every point on the perpendicular bisector is equidistant from both 𝐴 and 𝐵, the perpendicular bisector of a line is also its axis of symmetry.

Finally, we can also show that all points equidistant from 𝐴 and 𝐵 lie on the perpendicular bisector. If we let 𝐸 be any point that is equidistant from 𝐴 and 𝐵, then either 𝐸 is coincident with 𝐶 or triangles 𝐴𝐶𝐸 and 𝐵𝐶𝐸 are congruent by the SAS criterion.

Let’s now see an example of determining which quadrilateral is given by a geometric construction.

Example 4: Linking the Construction of the Perpendicular Bisector of a Line Segment with the Properties of a Rhombus

Draw the given figure and connect the points 𝐴𝐶𝐵𝐷. What does that figure represent?

Answer

We first note that 𝐶 and 𝐷 are the points of intersection between two circles of radius 𝐴𝐵 centered at 𝐴 and 𝐵. We recall that this means that 𝐶𝐷 is the perpendicular bisector of 𝐴𝐵. If we then add these right angles and quadrilateral 𝐴𝐶𝐵𝐷 onto the diagram and label the center point 𝐸, we get the following.

We note that 𝐴𝐸𝐶, 𝐵𝐸𝐶, 𝐷𝐵𝐸, and 𝐷𝐴𝐸 all have two congruent sides and the included angles are all right angles. So, all four triangles are congruent.

Thus, their corresponding sides are all congruent. This means all four sides of the quadrilateral are equal in length. We note that this is not necessarily a square, since we do not know if 𝐴𝐸 and 𝐶𝐸 have the same length. Hence, we can conclude that the quadrilateral is a rhombus.

Before we move on to our final example, we are equipped to discuss the concept of the shortest distance between a point and a line. Let’s say we have a line 𝐴𝐵 and a point 𝐶 and we want to determine the shortest distance from the point to the line.

We can first note that we can find the distance between any point on the line, 𝐷, and 𝐶 by finding the length of 𝐶𝐷.

However, if we also sketch the perpendicular line from 𝐴𝐵 to 𝐶, we can note that 𝐶𝐷 is the hypotenuse of a right triangle and so is longer than the perpendicular distance.

Hence, the perpendicular distance between a point and a line is the shortest distance.

In our final example, we will use this property to identify a geometric construction of the shortest distance between a point and a line.

Example 5: Identifying the Shortest Distance from a Point to a Line

In the following figure, what is the shortest distance between point 𝐷 and 𝐴𝐵?

Answer

We start by recalling that the shortest distance between a point and a line is the perpendicular distance. Thus, we need to find the perpendicular distance from point 𝐷 to 𝐴𝐵.

We note that the line 𝐴𝐵 is intersected by two circular arcs from a circle centered at 𝐷. Therefore, if we call the points of intersection between the line 𝐴𝐵 and this circle 𝐴 and 𝐵, then we know that 𝐴𝐷=𝐵𝐷, since these are radii of the same circle; it means that 𝐷 is equidistant from 𝐴 and 𝐵.

We then note that 𝐶 lies on circular arcs of the same radius centered at 𝐴 and 𝐵. So, 𝐴𝐶=𝐵𝐶, which means that 𝐶 is equidistant from 𝐴 and 𝐵.

Hence, both 𝐶 and 𝐷 are equidistant from 𝐴 and 𝐵, so they both lie on the perpendicular bisector of 𝐴𝐵. Therefore, the point of intersection between 𝐴𝐵 and 𝐶𝐷 is the perpendicular bisector of 𝐴𝐵; this is labeled 𝐸.

Thus, 𝐷𝐸𝐴𝐵 and 𝐸𝐴𝐵, so 𝐷𝐸 is the shortest distance from 𝐴𝐵 to 𝐷.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • To find the perpendicular line to 𝐴𝐵 passing through 𝐶𝐴𝐵, we use the following steps:
    1. We start by setting the point of our compass at 𝐶 and then trace a circle intersecting 𝐴𝐵 twice. We can label these points 𝐴 and 𝐵.
    2. We then set the radius of the compass to be larger than 𝐴𝐶, and we can trace circles centered at 𝐵 and 𝐴 such that they intersect at a point we will call 𝐷.
    3. We then have that 𝐶𝐷𝐴𝐵.
  • The perpendicular bisector of a line segment is the line that separates the line segment into two equal parts and meets the line segment at right angles. Every point on the perpendicular bisector of 𝐴𝐵 is equidistant from 𝐴 and 𝐵. All points equidistant from 𝐴 and 𝐵 lie on the perpendicular bisector of 𝐴𝐵. The perpendicular bisector of a line segment is its axis of symmetry.
  • To find the perpendicular bisector of 𝐴𝐵, we use the following steps:
    1. Set the radius of the compass to be greater than 𝐴𝐵.
    2. Sketch circles of this radius centered at 𝐴 and 𝐵. These will intersect at two points that we will name 𝐶 and 𝐷.
    3. 𝐶𝐷 is then the perpendicular bisector of 𝐴𝐵.
  • To find the perpendicular to 𝐴𝐵 passing through 𝐶𝐴𝐵, we use the following steps:
    1. Trace a circle centered at 𝐶 that intersects 𝐴𝐵 twice; label these points 𝐴 and 𝐵.
    2. Trace circles centered at 𝐴 and 𝐵 that intersect; label the point of intersection 𝐷.
    3. 𝐶𝐷 is then perpendicular to 𝐴𝐵.
  • The shortest distance between a line and a point is the perpendicular distance.

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