Question

If the point (a, 2a) falls between the lines |x + y + 1| = 4, then complete set of values of 'a' is

• A. $(-\frac{5}{3}, -1)$
• B. $(\frac{-5}{3}, \frac{1}{3})$
• C. $(\frac{-4}{3}, \frac{4}{3})$
• D. $(-\infty, \frac{-5}{3}) \cup (1, \infty)$

Answer 1

$(-\frac{5}{3}, 1)$

Answer 2

The problem asks for the complete set of values of 'a' such that the point $(a, 2a)$ lies between the lines given by the equation $|x + y + 1| = 4$.

First, let's break down the equation of the lines: $|x + y + 1| = 4$ implies two separate linear equations:

1. $x + y + 1 = 4 \Rightarrow x + y - 3 = 0$
2. $x + y + 1 = -4 \Rightarrow x + y + 5 = 0$

Let these lines be $L_1: x + y - 3 = 0$ and $L_2: x + y + 5 = 0$. These are two parallel lines.

For a point $(x_0, y_0)$ to fall "between" two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$, the expressions $(Ax_0 + By_0 + C_1)$ and $(Ax_0 + By_0 + C_2)$ must have opposite signs. This is equivalent to saying that their product must be negative: $(Ax_0 + By_0 + C_1)(Ax_0 + By_0 + C_2) < 0$.

Alternatively, and more directly for an absolute value inequality, the region between the lines $|X| = K$ (where $X = x+y+1$ and $K=4$) is given by $-K < X < K$. So, the condition for the point $(x, y)$ to be between the lines is: $-4 < x + y + 1 < 4$

Now, substitute the coordinates of the given point $(a, 2a)$ into this inequality: $-4 < a + 2a + 1 < 4$ $-4 < 3a + 1 < 4$

To solve this compound inequality, we can split it into two separate inequalities:

1. $3a + 1 < 4$
2. $3a + 1 > -4$

Let's solve each inequality:

1. $3a + 1 < 4$

$3a < 4 - 1$

$3a < 3$

$a < 1$ 2. $3a + 1 > -4$

$3a > -4 - 1$

$3a > -5$

$a > -\frac{5}{3}$

Combining the two conditions, $a < 1$ and $a > -\frac{5}{3}$, we get the complete set of values for 'a': $-\frac{5}{3} < a < 1$

In interval notation, this is $(-\frac{5}{3}, 1)$.

The calculated correct interval is $(-\frac{5}{3}, 1)$.