Understanding the Problem: Point Between Lines

The question asks for the condition under which a specific point, (a, a), lies in the region between two lines defined by the equation \(|x + y| = 2\). To solve this, we first need to understand what the equation \(|x + y| = 2\) represents and what it means for a point to lie between these lines.

Breaking Down the Equation \(|x + y| = 2\)

The absolute value equation \(|x + y| = 2\) can be split into two separate linear equations:

1. \(x + y = 2\)
2. \(x + y = -2\)

These two equations represent two distinct lines. Notice that both lines have a slope of -1 (\(y = -x + 2\) and \(y = -x - 2\)). Since they have the same slope, they are parallel lines.

Condition for a Point Lying Between Parallel Lines

A point \((x_0, y_0)\) lies between the parallel lines \(x + y = 2\) and \(x + y = -2\) if and only if the value of \(x_0 + y_0\) is strictly between -2 and 2. In mathematical terms, this condition is written as:

\(-2 < x_0 + y_0 < 2\)

Applying the Condition to Point (a, a)

The given point is (a, a). This means we substitute \(x_0 = a\) and \(y_0 = a\) into the inequality for the region between the lines:

\(-2 < a + a < 2\)

Simplify the expression in the middle:

\(-2 < 2a < 2\)

Solving for 'a'

To isolate 'a', we divide all parts of the inequality by 2. Since we are dividing by a positive number (2), the direction of the inequality signs does not change:

\(\frac{-2}{2} < \frac{2a}{2} < \frac{2}{2}\)

\(-1 < a < 1\)

Expressing the Result Using Absolute Value

The inequality \(-1 < a < 1\) means that the value of 'a' is between -1 and 1, not including -1 or 1. This is precisely the definition of \(|a| < 1\). The absolute value of 'a' is less than 1.

Comparing with Options

We found that the condition for the point (a, a) to lie between the lines \(|x + y| = 2\) is \(|a| < 1\). Let's look at the given options:

1. \(|a| < 2\)
2. \(|a| < \sqrt{2}\)
3. \(|a| < 1\)
4. \(|a| < \frac{1}{\sqrt{2}}\)

Our derived condition matches option 3.

Therefore, for the point (a, a) to lie between the lines \(|x + y| = 2\), the value of 'a' must satisfy \(|a| < 1\).

Revision Table: Key Concepts

Additional Information: Distance Between Parallel Lines

The two lines \(x+y=2\) and \(x+y=-2\) are parallel. The general form of a line is \(Ax + By + C = 0\). So, we have \(x+y-2=0\) and \(x+y+2=0\). Here, \(A=1\), \(B=1\). For the first line, \(C_1=-2\), and for the second line, \(C_2=2\).

The distance between two parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is given by the formula:

\(d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}\)

For our lines:

\(d = \frac{|-2 - 2|}{\sqrt{1^2 + 1^2}} = \frac{|-4|}{\sqrt{1 + 1}} = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}\)

The region between the lines is a strip of width \(2\sqrt{2}\). The midpoint of this region is the line \(x+y=0\). The point (a, a) lies on the line \(y=x\). We found that (a, a) is between the lines when \(-1 < a < 1\). This corresponds to points on the line \(y=x\) within the segment from (-1, -1) to (1, 1). The distance of point (a,a) from the line \(x+y=2\) is \(\frac{|a+a-2|}{\sqrt{1^2+1^2}} = \frac{|2a-2|}{\sqrt{2}}\). The distance of point (a,a) from the line \(x+y=-2\) is \(\frac{|a+a+2|}{\sqrt{1^2+1^2}} = \frac{|2a+2|}{\sqrt{2}}\). For the point to be between the lines, it must be on the same side of \(x+y=2\) as the origin (0,0), and on the same side of \(x+y=-2\) as the origin (0,0). For (0,0), \(0+0=0\), which is less than 2 and greater than -2. So, for (a,a), we need \(a+a < 2\) and \(a+a > -2\), which is exactly \(-2 < 2a < 2\), leading back to \(|a| < 1\).