# 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.

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.

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:

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:

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.

$x^2 - 5x + 4 = 0, \Rightarrow \left({x - 1}\right)\left({x - 4}\right) = 0, \Rightarrow x_1=1, x_2 = 4.$

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