The equation of the straight line which bisects the intercepts between the axes of the lines `x+y=2 `and `2x + 3y = 6` is

Options

A. `2x=3`
B. `y=1`
C. `2y=3`
D. `x=1`

Answer

To find the equation of the straight line that bisects the intercepts between the axes of the lines \(x + y = 2\) and \(2x + 3y = 6\), we will follow these steps:

### Step 1: Find the intercepts of the first line \(x + y = 2\)

To find the x-intercept, set \(y = 0\):
\[ x + 0 = 2 \implies x = 2 \]
So, the x-intercept is \((2, 0)\).

To find the y-intercept, set \(x = 0\):
\[ 0 + y = 2 \implies y = 2 \]
So, the y-intercept is \((0, 2)\).

### Step 2: Find the intercepts of the second line \(2x + 3y = 6\)

To find the x-intercept, set \(y = 0\):
\[ 2x + 0 = 6 \implies 2x = 6 \implies x = 3 \]
So, the x-intercept is \((3, 0)\).

To find the y-intercept, set \(x = 0\):
\[ 0 + 3y = 6 \implies 3y = 6 \implies y = 2 \]
So, the y-intercept is \((0, 2)\).

### Step 3: Identify the intercept points

From the calculations:
- The intercepts of the first line \(x + y = 2\) are \((2, 0)\) and \((0, 2)\).
- The intercepts of the second line \(2x + 3y = 6\) are \((3, 0)\) and \((0, 2)\).

### Step 4: Find the midpoints of the intercepts

For the first line, the midpoint \(P_1\) of the intercepts \((2, 0)\) and \((0, 2)\) is calculated as:
\[ P_1 = \left(\frac{2 + 0}{2}, \frac{0 + 2}{2}\right) = \left(1, 1\right) \]

For the second line, the midpoint \(P_2\) of the intercepts \((3, 0)\) and \((0, 2)\) is calculated as:
\[ P_2 = \left(\frac{3 + 0}{2}, \frac{0 + 2}{2}\right) = \left(\frac{3}{2}, 1\right) \]

### Step 5: Find the equation of the line passing through the midpoints \(P_1\) and \(P_2\)

The coordinates of \(P_1\) are \((1, 1)\) and the coordinates of \(P_2\) are \(\left(\frac{3}{2}, 1\right)\).

The slope \(m\) of the line through these points is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 1}{\frac{3}{2} - 1} = \frac{0}{\frac{1}{2}} = 0 \]

Since the slope is \(0\), the line is horizontal.

### Step 6: Write the equation of the line

The equation of a horizontal line passing through \(y = 1\) is:
\[ y = 1 \]

### Final Answer

Thus, the equation of the straight line that bisects the intercepts is:
\[ \boxed{y = 1} \]