Precalculus

Analytic Geometry

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Cartesian Coordinates on a Line

Position of a Point on a Straight Line

Cartesian coordinates for a one-dimensional space are defined as follows. Take a straight line and choose an arbitrary point O as the origin. Specify a unit of measure which allows us to measure the lengths of segments of the line.

To uniquely determine the x-coordinate of any point M, we also need to specify a certain direction of the line. It is usually assumed that x > 0 the point M is located to the right of the origin O and, accordingly, x < 0 if the point M is located to the left. We will mark the positive direction of the coordinate line with an arrow.

One-dimensional coordinate system
Figure 1.

The fact that point M has coordinate x is denoted as \(x_M,\) \(x_1\) or \({M\left({x}\right),}\) etc. The origin O has coordinate \(x = 0.\)

By introducing Cartesian coordinates on the line, any point \(M\) on the straight line is associated with a well-defined real number \(x.\) There is a one-to-one correspondence between the real numbers and points on a line.

The one-dimensional coordinate system introduced in this way is also known as a number line.

Displacement

Consider two arbitrary points M1 and M2 with coordinates x1, x2, respectively.

Displacement between two points on a straight line
Figure 2.

The displacement of point M2 with respect to point M1 in a given one-dimensional coordinate system is determined by the formula

\[\Delta x = x_2 - x_1.\]

Distance Between Two Points

The distance \(d\left({M_1,M_2}\right)\) between points \(M\left({x_1}\right)\) and \(M\left({x_2}\right)\) on the coordinate line is defined by the formula

\[d\left({M_1,M_2}\right) = \left|{\Delta x}\right| = \left|{x_2 - x_1}\right|.\]

Center of Gravity of a System of Material Points

Let points \(M_1, M_2,\ldots,M_n\) have coordinates \(x_1, x_2,\ldots,x_n\) and masses, respectively, \(m_1, m_2,\ldots,m_n.\) Then the coordinate of the center of gravity of such a system is calculated by the formula

\[x = \frac{m_1x_1 + m_2x_2 + \ldots + m_nx_n}{m_1 + m_2 + \ldots + m_n}.\]

Solved Problems

Click or tap a problem to see the solution.

Example 1

Given the coordinates of three points \(M_1\left({-3}\right),\) \(M_2\left({-7}\right),\) \(M_3\left({2}\right)\) on a line. Determine the displacement and distance for all pairs of points.

Example 2

Indicate on the number line all points whose coordinates \(x\) satisfy the inequality

\[1 \le x^2 \le 2.\]

Example 3

Find points on the number line whose coordinates \(x\) satisfy the equation

\[\left|{3 + x}\right| = 3.\]

Example 4

Which of the two points lies to the right: \(M\left({x}\right)\) or \(L\left({2x}\right)?\)

Example 5

Three material points with masses \(m_1 = 3, m_2 = 2, m_3 = 1\) are located on the number line and have coordinates \(x_1 = 1,\) \(x_2 = 2,\) \(x_3 = 3,\) respectively. Find the coordinate of the center of gravity of the system.

Example 6

Indicate on the number line the set of points whose coordinates satisfy the inequality

\[\left|{x^2 - 5x + 4}\right| \gt x^2 - 5x + 4.\]

Example 1.

Given the coordinates of three points \(M_1\left({-3}\right),\) \(M_2\left({-7}\right),\) \(M_3\left({2}\right)\) on a line. Determine the displacement and distance for all pairs of points.

Solution.

The figure below shows how the given points are located on the straight line:

Three points on a straight line
Figure 3.

The displacement is a signed quantity. Therefore we have \(6\) possible values:

\[\Delta_{12} = x_2 - x_1 = -7 - \left({-3}\right) = -4;\]
\[\Delta_{21} = x_1 - x_2 = -3 - \left({-7}\right) = 4;\]
\[\Delta_{13} = x_3 - x_1 = 2 - \left({-3}\right) = 5;\]
\[\Delta_{31} = x_1 - x_3 = -3 - 2 = -5;\]
\[\Delta_{23} = x_3 - x_2 = 2 - \left({-7}\right) = 9;\]
\[\Delta_{32} = x_2 - x_3 = -7 - 2 = -9.\]

The distance is defined as the absolute value of displacement. So we only have three distance values:

\[d_{12} = d_{21} = \left|\Delta_{12}\right| = \left|\Delta_{21}\right| = 4;\]
\[d_{13} = d_{31} = \left|\Delta_{13}\right| = \left|\Delta_{31}\right| = 5;\]
\[d_{23} = d_{32} = \left|\Delta_{23}\right| = \left|\Delta_{32}\right| = 9.\]

Example 2.

Indicate on the number line all points whose coordinates \(x\) satisfy the inequality

\[1 \le x^2 \le 2.\]

Solution.

This double inequality has the following solution:

\[1 \le x^2 \le 2, \Rightarrow \left[\begin{array}{l} -\sqrt{2} \le x \le -1\\ 1 \le x \le \sqrt{2} \end{array} \right..\]

On the coordinate axis, the solution is depicted as two segments:

Solution of the inequality 1<=x^2<=2
Figure 4.

The answer is written as the set

\[x \in \left[{-\sqrt{2},-1}\right] \cup \left[{1,\sqrt{2}}\right].\]

Example 3.

Find points on the number line whose coordinates \(x\) satisfy the equation

\[\left|{3 + x}\right| = 3.\]

Solution.

We replace the absolute value equation with two equivalent equations:

\[\left|{3 + x}\right| = 3, \Rightarrow \left[\begin{array}{l} -\left({3 + x}\right) = 3 &\text{if} &3 + x \lt 0\\ 3 + x = 3, &\text{if} &3 + x \ge 0 \end{array} \right..\]

Solve both these equations:

\[\Rightarrow \left[\begin{array}{l} -3 - x = 3 &\text{if} &x \lt -3\\ 3 + x = 3, &\text{if} &x \ge -3 \end{array} \right., \Rightarrow \left[\begin{array}{l} x = -6 &\text{if} &x \lt -3\\ x = 0, &\text{if} &x \ge -3 \end{array} \right., \Rightarrow \left[\begin{array}{l} x = -6\\ x = 0 \end{array} \right..\]

Thus there are two points \(x = -6,\) \(x = 0\) that satisfy the given equation.

Example 4.

Which of the two points lies to the right: \(M\left({x}\right)\) or \(L\left({2x}\right)?\)

Solution.

If \(x \gt 0,\) then

\[x + x \gt 0 + x, \Rightarrow 2x \gt x,\]

that is, the point \(L\) lies to the right of the point \(M.\)

If \(x \lt 0,\) then

\[x + x \lt 0 + x, \Rightarrow 2x \lt x,\]

that is, the point \(L\) lies to the left of the point \(M.\)

If \(x = 0,\) then

\[2x = 0.\]

In this case both points coincide.

Example 5.

Three material points with masses \(m_1 = 3, m_2 = 2, m_3 = 1\) are located on the number line and have coordinates \(x_1 = 1,\) \(x_2 = 2,\) \(x_3 = 3,\) respectively. Find the coordinate of the center of gravity of the system.

Solution.

Calculate the coordinate \(x\) of the center of gravity using the formula

\[x = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}.\]

Substitute known values:

\[x = \frac{3\cdot1 + 2\cdot2 + 1\cdot3}{3 + 2 + 1} = \frac{10}{6} = \frac{5}{3}.\]

Example 6.

Indicate on the number line the set of points whose coordinates satisfy the inequality

\[\left|{x^2 - 5x + 4}\right| \gt x^2 - 5x + 4.\]

Solution.

We first solve the quadratic equation and plot the graph of the function.

\[x^2 - 5x + 4 = 0, \Rightarrow \left({x - 1}\right)\left({x - 4}\right) = 0, \Rightarrow x_1=1, x_2 = 4.\]
Graph of the quadratic function y=x^2-5x+4
Figure 5.

To get rid of the absolute value consider two cases.

Case 1.

If \(x \in \left({-\infty, 1}\right] \cup \left[{4, \infty}\right),\) then \(x^2 - 5x + 4 \ge 0\) and the inequality takes the form

\[x^2 - 5x + 4 \gt x^2 - 5x + 4, \Rightarrow 0 \gt 0.\]

Obviously this inequality has no solutions: \(x \in \varnothing.\)

Case 2.

If \(x \in \left({1, 4}\right),\) then \(x^2 - 5x + 4 \lt 0\) and the inequality is written as

\[-\left({x^2 - 5x + 4}\right) \gt x^2 - 5x + 4, \Rightarrow 2\left({x^2 - 5x + 4}\right) \lt 0, \Rightarrow x^2 - 5x + 4 \lt 0, \Rightarrow x \in \left({1, 4}\right).\]

Answer: \(x \in \left({1, 4}\right).\)