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Let \(P\) be a point and \(m\) a line not passing through \(P\). Prove that, among all possible line segments \(\underline{P X}\) with \(X\) on the line \(m\), a perpendicular from \(P\) to the line \(m\) is the shortest.

Short Answer

Expert verified
The perpendicular from \( P \) to line \( m \) is the shortest line.

Step by step solution

01

Define the Problem

We need to show that the shortest distance from a point \( P \) to a line \( m \) is along the perpendicular drawn from \( P \) to \( m \). This means that for any point \( X \) on line \( m \), the segment \( \underline{PX} \) is shortest when \( PX \) is perpendicular to \( m \).
02

Understand the Geometry

Visualize the geometry: Assume \( m \) is a horizontal line, and \( P \) is a point above it. Draw different lines from \( P \) to various points \( X \) on \( m \). Notice that one of these lines will be perpendicular to \( m \).
03

Apply Triangle Inequality Theorem

For a point \( X \) not directly below \( P \), create a right triangle with vertices \( P, X, \) and a perpendicular \( Y \), where \( Y \) is the foot of the perpendicular from \( P \) to \( m \). By the triangle inequality theorem, \( PY < PX \).
04

Conclude with Logic

Since the hypotenuse \( PX \) of any right triangle \( \triangle PXY \) is longer than the leg \( PY \), the line segment \( PY \) (the perpendicular) is the shortest line from \( P \) to any point on \( m \). Thus, the perpendicular from \( P \) to \( m \) is indeed the shortest distance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triangle Inequality Theorem
The Triangle Inequality Theorem is a fundamental principle in geometry. It states that in any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. This theorem is particularly useful when assessing the shortest paths and distances in geometric problems.

In the context of the problem at hand, consider a triangle formed by a point \( P \), a point \( X \) on the line \( m \), and \( Y \), which is the point where a perpendicular from \( P \) meets the line \( m \). The triangle \( \triangle PXY \) illustrates this principle beautifully. Here, the line segment \( PX \) is the hypotenuse, while \( PY \) is opposite the right angle.

According to the Triangle Inequality Theorem:
  • The hypotenuse \( PX \) will always be longer than the perpendicular leg \( PY \).
  • This leads to the conclusion that the perpendicular distance \( PY \) from \( P \) to the line \( m \) is the shortest possible distance.
This understanding highlights how the Triangle Inequality Theorem helps us conclude why the perpendicular segment is the minimal distance between a point and a line.
Geometry Visualization
Visualizing geometric concepts is integral to understanding and solving geometry problems, especially in proving theorems and properties. With practice, the ability to imagine shapes, angles, and lines develops over time and becomes an invaluable tool in mathematics.

In this exercise, visualizing the relationship between the point \( P \) and the line \( m \) helps in understanding why the perpendicular line segment \( PY \) is the shortest distance. Imagine line \( m \) as horizontal, and point \( P \) positioned above it. From \( P \), you can draw numerous line segments intersecting at different points along line \( m \).

Among all these potential line segments, visualizing that there's one line segment at a 90-degree angle is crucial. This specific line is the perpendicular, and because of its perpendicularity, it represents the shortest distance from \( P \) to line \( m \).
  • Visualize different orientations and confirm which one ensures perpendicularity.
  • Observe that as the angle approaches 90 degrees with the line, the length of the segment is minimized.
Through this visualization, understanding how a perpendicular from a point to a line creates the shortest path becomes evident.
Right Triangle Properties
A right triangle has specific properties that make it a critical shape in geometry. One of these properties is that it has one right angle (90 degrees), and the side opposite this angle is the longest, called the hypotenuse.

In our problem, when we drop a perpendicular \( PY \) from point \( P \) to line \( m \), we effectively create a right triangle \( \triangle PXY \). Here's why the properties of a right triangle matter:
  • The hypotenuse (\( PX \)) is always longer than either of its legs (\( PY \) and \( XY \)).
  • Among all triangles that include a given line segment, a right triangle will have the shortest possible leg opposite the right angle.
Therefore, by employing right triangle properties, we understand that the perpendicular leg \( PY \) from \( P \) to \( m \) is shorter than any potential hypotenuse \( PX \) that we might draw to any other \( X \) on the line \( m \). This intrinsic property of right triangles reinforces the conclusion that the perpendicular is the shortest distance between a point and a line.

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Most popular questions from this chapter

Suppose that in \(\triangle A B C, \angle C=\angle A+\angle B\). Prove that \(C\) lies on the circle with diameter \(\underline{A B}\). (In particular, if the angles of \(\triangle A B C\) add to a straight angle, and \(\angle A C B\) is a right angle, then \(C\) lies on the circle with diameter \(\underline{A B}\).)

(a) Find a formula for the surface area of a right cylinder with height \(h\) and with circular base of radius \(r\). (b) Find a similar formula for the surface area of a right prism with height \(h\), whose base is a regular \(n\) -gon with inradius \(r\).

The only possible path along the edges of a 2 D-cube uses each vertex once and returns to the start after visiting all four vertices. (a)(i) Draw a path along the edges of a 3D-cube that visits each vertex exactly once and returns to the start. (ii) Look at the sequence of coordinate triples as you follow your path. What do you notice? (b)(i) Draw a path along the edges of a \(4 \mathrm{D}\) -cube that visits each vertex exactly once and returns to the start. (ii) Look at the sequence of coordinate 4 -tuples as you follow your path. What do you notice?

A vertex figure is to be formed by fitting regular \(p\) -gons together, edge- to-edge, for a fixed \(p .\) If there are \(q\) of these \(p\) -gons at a vertex, we denote the vertex figure by \(p^{q} .\) If the angles at each vertex add to less than \(360^{\circ}\), prove that the only possible vertex figures are \(3^{3}, 3^{4}, 3^{5}\), \(4^{3}, 5^{3}\)

(a) In the quadrilateral \(A B C D\) the two diagonals \(\underline{A C}\) and \(\underline{B D}\) cross at \(X\). Suppose \(\underline{A B}=\underline{B C}, \angle B A C=60^{\circ}, \angle D A C=40^{\circ}, \angle B X C=100^{\circ}\) (i) Calculate (exactly) \(\angle A D B\) and \(\angle C B D\). (ii) Calculate \(\angle B D C\) and \(\angle A C D\). (b) In the quadrilateral \(A B C D\) the two diagonals \(\underline{A C}\) and \(\underline{B D}\) cross at \(X\). Suppose \(\underline{A B}=\underline{B C}, \angle B A C=70^{\circ}, \angle D A C=40^{\circ}, \angle B X C=100^{\circ}\) (i) Calculate (exactly) the size of \(\angle B D C+\angle A C D\) (ii) Explain how we can be sure that \(\angle B D C\) and \(\angle A C D\) are uniquely determined, even though we cannot calculate them immediately. \(\triangle\)

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