If `(x , y)` is a variable point on the line `y=2x` lying between the lines `2(x+1)+y=0` and `x+3(y-1)=0` , then `x in (-1/2,6/7)` (b) `x in (-1/2,3/7)` `y in (-1,3/7)` (d) `y in (-1,6/7)`

Options

A. `x in (-1//2, 6//7)`
B. `x in (-1//2, 3//7)`
C. `y in (-1, 3//7)`
D. `y in (-1, 6//7)`

Explanation

To find the range of \(x\) and \(y\) for the variable point \((x, y)\) on the line \(y = 2x\) that lies between the lines \(2(x + 1) + y = 0\) and \(x + 3(y - 1) = 0\), we will follow these steps:

### Step 1: Write down the equations of the lines.
The equations given are:
1. \(y = 2x\) (the line on which the point lies)
2. \(2(x + 1) + y = 0\) (let's call this Line 1)
3. \(x + 3(y - 1) = 0\) (let's call this Line 2)

### Step 2: Convert Line 1 into slope-intercept form.
Starting with Line 1:
\[ 2(x + 1) + y = 0 \]
Expanding this gives:
\[ 2x + 2 + y = 0 \implies y = -2x - 2 \]

### Step 3: Convert Line 2 into slope-intercept form.
Starting with Line 2:
\[ x + 3(y - 1) = 0 \]
Expanding this gives:
\[ x + 3y - 3 = 0 \implies 3y = -x + 3 \implies y = -\frac{1}{3}x + 1 \]

### Step 4: Find the intersection points of the lines.
#### Intersection of \(y = 2x\) and Line 1:
Substituting \(y = 2x\) into Line 1:
\[ 2x = -2x - 2 \]
Combining like terms:
\[ 4x = -2 \implies x = -\frac{1}{2} \]
Now, substituting \(x = -\frac{1}{2}\) back into \(y = 2x\):
\[ y = 2\left(-\frac{1}{2}\right) = -1 \]
So, the intersection point is \(\left(-\frac{1}{2}, -1\right)\).

#### Intersection of \(y = 2x\) and Line 2:
Substituting \(y = 2x\) into Line 2:
\[ 2x = -\frac{1}{3}x + 1 \]
Multiplying through by 3 to eliminate the fraction:
\[ 6x = -x + 3 \]
Combining like terms:
\[ 7x = 3 \implies x = \frac{3}{7} \]
Now, substituting \(x = \frac{3}{7}\) back into \(y = 2x\):
\[ y = 2\left(\frac{3}{7}\right) = \frac{6}{7} \]
So, the intersection point is \(\left(\frac{3}{7}, \frac{6}{7}\right)\).

### Step 5: Determine the ranges for \(x\) and \(y\).
From the intersection points, we have:
- Point A: \(\left(-\frac{1}{2}, -1\right)\)
- Point B: \(\left(\frac{3}{7}, \frac{6}{7}\right)\)

Thus, the ranges for \(x\) and \(y\) are:
- For \(x\): \(x \in \left(-\frac{1}{2}, \frac{3}{7}\right)\)
- For \(y\): \(y \in \left(-1, \frac{6}{7}\right)\)

### Final Answer:
- \(x \in \left(-\frac{1}{2}, \frac{3}{7}\right)\)
- \(y \in \left(-1, \frac{6}{7}\right)\)