If the point `P((3a)/(2),1)` lies between the two different lines `x+y=a` and `x+y=2a`, then the least integral value of `|a|` is equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the least integral value of \(|a|\) such that the point \(P\left(\frac{3a}{2}, 1\right)\) lies between the lines \(x + y = a\) and \(x + y = 2a\), we can follow these steps:

### Step 1: Understand the Position of the Point
We need to determine the position of the point \(P\) relative to the two lines. The lines are given as:
1. \(L_1: x + y = a\)
2. \(L_2: x + y = 2a\)

### Step 2: Substitute the Point into the Line Equations
We will substitute the coordinates of point \(P\) into the equations of the lines to find the inequalities that must hold for \(P\) to lie between the two lines.

1. For line \(L_1\):
\[ \frac{3a}{2} + 1 > a \]
Rearranging gives:
\[ \frac{3a}{2} + 1 - a > 0 \implies \frac{3a}{2} - \frac{2a}{2} + 1 > 0 \implies \frac{a}{2} + 1 > 0 \implies a > -2 \]

2. For line \(L_2\):
\[ \frac{3a}{2} + 1 < 2a \]
Rearranging gives:
\[ \frac{3a}{2} + 1 - 2a < 0 \implies \frac{3a}{2} - \frac{4a}{2} + 1 < 0 \implies -\frac{a}{2} + 1 < 0 \implies -a + 2 < 0 \implies a > 2 \]

### Step 3: Combine the Inequalities
From the inequalities derived:
1. \(a > -2\)
2. \(a > 2\)

Since \(a > 2\) is the stronger condition, we will focus on this inequality.

### Step 4: Find the Least Integral Value of \(|a|\)
The least integral value of \(a\) that satisfies \(a > 2\) is \(3\). Therefore, we have:
\[ |a| = 3 \]

### Conclusion
Thus, the least integral value of \(|a|\) is:
\[ \boxed{3} \]