Lesson Video: Geometric Constructions: Perpendicular Lines Mathematics

In this video, 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.

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Video Transcript

In this video, 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 note that for any line 𝐴𝐡, there is a unique line that is perpendicular to line 𝐴𝐡 that passes through a given point 𝐢 on line 𝐴𝐡. We can construct this line with a compass and a straightedge by using our knowledge of circles and triangles. We start by setting the point of our compass at 𝐢. Then, we can use any size for the radius. There is no need to trace the whole circle. So we only show the arcs where the circle intersects the line.

We call the points of intersection of the circle and line 𝐴𝐡 𝐡 prime and 𝐴 prime, respectively. We note that 𝐡 prime 𝐢 equals 𝐴 prime 𝐢, since these are both radii of the same circle centered on point 𝐢. So we can say that 𝐢 is equidistant from 𝐴 prime and 𝐡 prime.

Next, we will set the radius of our compass to be larger than 𝐴 prime 𝐢 and construct a pair of congruent circles centered at 𝐡 prime and 𝐴 prime. Instead of constructing full circles, we can just sketch the arcs where the circles intersect, either above or below line 𝐴𝐡. In this case, we call the point of intersection above line 𝐴𝐡 point 𝐷.

We note that 𝐡 prime 𝐷 equals 𝐴 prime 𝐷, as they are radii of congruent circles, one centered on point 𝐡 prime and the other centered on point 𝐴 prime. Finally, we use our straightedge to connect points 𝐢 and 𝐷 with a line segment. By the reflexive property, 𝐷𝐢 equals 𝐷𝐢. We note that triangles 𝐡 prime 𝐢𝐷 and 𝐴 prime 𝐢𝐷 have congruent sides. So by the side-side-side triangle congruence criterion, these are congruent triangles. Then, because corresponding parts of congruent triangles are congruent, it follows that angle 𝐴 prime 𝐢𝐷 is congruent to angle 𝐡 prime 𝐢𝐷.

For two congruent angles to make up a straight angle, as we see here, their angles must sum to 180 degrees. And in this case, since the two angles are congruent, they must both be right angles. This proves that 𝐢𝐷 is a unique line perpendicular to line 𝐴𝐡 passing through point 𝐢.

Now we will clear a bit of space in order to summarize these steps we used to construct the perpendicular.

To find the perpendicular line to 𝐴𝐡 passing through point 𝐢 on line 𝐴𝐡, we use the following steps. We start by setting the point of our compass at 𝐢, then tracing a circle intersecting line 𝐴𝐡 twice. We label these points 𝐡 prime and 𝐴 prime. Then, we set the radius of the compass to be larger than 𝐴 prime 𝐢, and we trace arcs of congruent circles centered at 𝐡 prime and 𝐴 prime such that they intersect. We call the point of intersection 𝐷. Finally, we have that line 𝐢𝐷 is perpendicular to line 𝐴𝐡. This technique allows us to find the perpendicular line passing through any given point on a line.

Next, we will look at a special case of this that splits a segment into two equal parts called the perpendicular bisector.

In general, to bisect a line segment is to separate it into two equal parts, in other words to cross through the midpoint of the line segment. In particular, the perpendicular bisector of a line segment is the line that separates the segment into two equal parts and meets the segment at right angles. We will use our compass and straightedge again to construct the perpendicular bisector of any line segment.

To find the perpendicular bisector to line segment 𝐴𝐡, we use the following steps. We begin by setting the radius of the compass to be greater than half the length of 𝐴𝐡. Then, we trace two arcs of the circle of this radius centered at 𝐴 and two intersecting arcs of a congruent circle centered at 𝐡. These arcs will intersect at two points that we label 𝐢 and 𝐷.

We then connect 𝐢 and 𝐷 with a straight line. We claim that line 𝐢𝐷 is the perpendicular bisector of line segment 𝐴𝐡. But we still need to prove this is true. To show our reasoning, we use our straightedge to draw in the radii of the circles centered at 𝐴 and 𝐡. Because the radii are equal in length, it follows that triangle 𝐡𝐢𝐷 is congruent to triangle 𝐴𝐢𝐷 by the side-side-side triangle congruence criterion. We know that their corresponding angles must be congruent as well, so we have angle 𝐡𝐢𝐷 congruent to angle 𝐴𝐢𝐷.

We label the intersection of 𝐢𝐷 and 𝐴𝐡 point 𝐸. We see that triangle 𝐴𝐸𝐢 and triangle 𝐡𝐸𝐢 share two congruent sides and the included angles are congruent. Thus, triangle 𝐴𝐸𝐢 is congruent to triangle 𝐡𝐸𝐢 by the side-angle-side triangle congruence criterion.

We know that their corresponding sides must be congruent, so we have 𝐴𝐸 equal to 𝐸𝐡. We also note that angle 𝐴𝐸𝐢 is congruent to angle 𝐡𝐸𝐢. Since these angles sum to make a straight angle, they must be right angles.

In summary, we have just shown that line 𝐢𝐷 separates line segment 𝐴𝐡 into two equal parts and meets 𝐴𝐡 at right angles. This means that line 𝐢𝐷 is the perpendicular bisector of line segment 𝐴𝐡.

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

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

Before connecting 𝐴𝐢𝐡𝐷, let’s label the intersection of line segments 𝐢𝐷 and 𝐴𝐡 point 𝐸. We see from the diagram that 𝐴𝐸 and 𝐸𝐡 have the same length and 𝐢𝐸 and 𝐸𝐷 have the same length. So we know that 𝐢𝐷 bisects line segment 𝐴𝐡 and 𝐴𝐡 bisects line segment 𝐢𝐷. We will now use a straightedge to connect the points 𝐴, 𝐢, 𝐡, and 𝐷 to form quadrilateral 𝐴𝐢𝐡𝐷.

To answer this question, we must look for key features of the quadrilateral to determine what kind of quadrilateral we are looking at.

We recall that a quadrilateral can be classified as a trapezoid or a kite or a parallelogram, such as a rectangle, rhombus, or square. We note that the intersecting arcs in the diagram represent the intersection between two circles. The orange circle is centered at 𝐴 with a radius 𝐴𝐢, and the pink circle is centered at 𝐡 with radius 𝐡𝐢. We recognize this construction as the perpendicular bisector construction.

This means that line 𝐢𝐷 not only bisects line segment 𝐴𝐡, but it is the perpendicular bisector of line segment 𝐴𝐡. Therefore, 𝐢𝐡 bisects 𝐴𝐡 at right angles, which we can add to the diagram. Therefore, the diagonals of 𝐴𝐢𝐡𝐷 are perpendicular. We know that the only quadrilaterals that have perpendicular diagonals are kites, rhombuses, and squares. Therefore, it is safe to eliminate trapezoids and rectangles from our list of possibilities.

Based on the accumulated evidence, it follows that triangles 𝐴𝐸𝐢, 𝐡𝐸𝐢, 𝐡𝐸𝐷, and 𝐴𝐸𝐷 are congruent by the side-angle-side triangle congruence criterion, since all four triangles have two congruent sides and the included angles are all right angles. Thus, the corresponding sides of the triangles are also congruent. This means all four sides of quadrilateral 𝐴𝐢𝐡𝐷 are equal in length.

We know that rhombuses have four congruent sides. And because a square is a specific type of rhombus, squares also have the same property. We can eliminate kites as a possibility, because they have two pairs of congruent sides. We recall that for a rhombus to be a square, the diagonals must be congruent. Since we do not know if 𝐴𝐸 and 𝐢𝐸 have the same length, we can’t prove that the diagonals are congruent. And therefore, we can’t claim that 𝐴𝐢𝐡𝐷 is a square.

Based on the evidence of the perpendicular diagonals and the congruent sides, we conclude that quadrilateral 𝐴𝐢𝐡𝐷 is a rhombus.

Before we move on to the next example, there is one more construction we need to consider.

We have already seen how to find the perpendicular line to 𝐴𝐡 through a point 𝐢 on the line. But what if 𝐢 does not lie on the line? To find the perpendicular to line 𝐴𝐡 passing through a point 𝐢 not on the line, we use the following steps. First, we place the point of our compass on 𝐢. Then we trace the arc of a circle that intersects line 𝐴𝐡 twice. We label these points 𝐡 prime and 𝐴 prime. We note that 𝐢 is equidistant from 𝐡 prime and 𝐴 prime, because 𝐢𝐡 prime and 𝐢𝐴 prime are radii of the same circle centered at 𝐢.

Next, we want to find another point equidistant from 𝐡 prime and 𝐴 prime below the line 𝐴𝐡. We do this by tracing the arcs of two intersecting circles centered at 𝐡 prime and 𝐴 prime using the same radius we started with. We name the point of intersection 𝐷. Since we kept the radii of the circles the same, 𝐡 prime 𝐷 equals 𝐴 prime 𝐷 equals 𝐡 prime 𝐢 equals 𝐴 prime 𝐢, which we add to the diagram. We recognize the resulting quadrilateral as a rhombus.

Next, we use our straightedge to connect vertex 𝐢 to vertex 𝐷 in the diagram. We recall that the diagonals of a rhombus are perpendicular. Therefore, line segment 𝐢𝐷 is perpendicular to line 𝐴𝐡. We can follow these same steps to construct the perpendicular line through any point not on the line. Let’s see if we can recognize one of these new constructions in our next example.

Which of the following constructions represents drawing a perpendicular from a point lying outside a straight line? We recall that to find a perpendicular from a point not on a line, we first trace a circle centered at that point which will intersect the line at the two distinct points. Then, we would trace circles with the same radius centered on the two points of intersection. These two circles would intersect at a point which we connect back to the original point, resulting in a line segment perpendicular to the original line.

Of the five diagrams given, the only one that matches these construction steps is option (E). First, we notice point 𝐢 is not on line 𝐴𝐡. We also see a circle centered on 𝐢 intersecting the line at points 𝐷 and 𝐹. Then, we notice the intersecting arcs of two circles centered on 𝐷 and 𝐹. This point of intersection then connects back to point 𝐢, forming a line perpendicular to line 𝐴𝐡.

It is worth noting that the other options give different geometric constructions. Option (A) is the construction of a perpendicular bisector. Option (B) finds the perpendicular 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 the construction of an angle bisector.

In our final example, we will consider how to find the shortest distance between a point and a line.

True or False: The shortest distance from a point to a line equals the length of any line segment that passes through the point and intersects with the line.

To answer this question, we’ll begin with a point and a line. Let’s the name the point 𝐢 and the line 𝐴𝐡. The statement provided is regarding the shortest distance from 𝐢 to line 𝐴𝐡. The claim is that any line segment through the point to the line is the shortest.

So let’s sketch an arbitrary line segment from the point to the line. Let’s name the point of intersection with the line point 𝐷. If the given statement is true, then we will not find any shorter distance from 𝐢 to line 𝐴𝐡. Let’s try sketching the perpendicular distance from 𝐢 to the line. We label the point 𝐸 where the perpendicular segment meets line 𝐴𝐡 at a right angle.

We notice that we have formed a right triangle: triangle 𝐢𝐸𝐷. Line segment 𝐢𝐷 is the hypotenuse of this right triangle. We recall that the hypotenuse of a right triangle is always the longest side. Therefore, 𝐢𝐷 is greater than 𝐢𝐸. This means that the given statement is false. Any nonperpendicular line segment from point 𝐢 to line 𝐴𝐡 would be considered a hypotenuse and thus be longer than the perpendicular distance.

We conclude that the perpendicular distance between a point and a line is the shortest distance.

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

We learned how to construct a perpendicular line through a point on the line. The first step is to trace a circle centered on point 𝐢 and label the points of intersection 𝐡 prime and 𝐴 prime. Then, we set a radius longer than 𝐴 prime 𝐢 and trace two intersecting circles centered on 𝐡 prime and 𝐴 prime. We label this intersection point 𝐷. The line passing through 𝐢 and 𝐷 is perpendicular to line 𝐴𝐡.

We also learned how to find the perpendicular to a line through a point not on the line. We begin by tracing a circle centered on point 𝐢, intersecting line 𝐴𝐡 at two points, which we name 𝐡 prime and 𝐴 prime. Next, we trace two circles with the same radius centered at 𝐡 prime and 𝐴 prime. We label their intersection below the line point 𝐷. Then, the line passing through points 𝐢 and 𝐷 is perpendicular to line 𝐴𝐡.

We also learned how to construct the perpendicular bisector of a line segment. A perpendicular bisector separates the line into two equal parts and meets the line segment at right angles. To begin, we set the radius of the compass to be greater than half the length of 𝐴𝐡. Then, we trace two circles of this radius centered at 𝐴 and 𝐡, intersecting both above and below the line segment 𝐴𝐡. We label these intersections 𝐢 and 𝐷. The line passing through 𝐢 and 𝐷 is the perpendicular bisector of line segment 𝐴𝐡. We note that all points on the perpendicular bisector are equidistant from both 𝐴 and 𝐡, since the distance is given by the same radius length.

And finally, we learned that the shortest distance between a line and a point is the perpendicular distance.

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