The equation of the straight line through the intersection of line `2x+y=1` and `3x+2y=5` passing through the origin, is:

Options

A. `7x+3y=0`
B. `7x-y=0`
C. `3x+2y=0`
D. `x+y=0`

Answer

To find the equation of the straight line that passes through the intersection of the lines \(2x + y = 1\) and \(3x + 2y = 5\) and also passes through the origin, we can follow these steps:

### Step 1: Find the intersection point of the two lines
We have the equations:
1. \(2x + y = 1\) (Equation 1)
2. \(3x + 2y = 5\) (Equation 2)

To find the intersection, we can solve these equations simultaneously.

### Step 2: Solve the equations
From Equation 1, we can express \(y\) in terms of \(x\):
\[ y = 1 - 2x \]

Now, substitute this expression for \(y\) into Equation 2:
\[ 3x + 2(1 - 2x) = 5 \]
Expanding this gives:
\[ 3x + 2 - 4x = 5 \]
Combining like terms:
\[ -1x + 2 = 5 \]
Subtracting 2 from both sides:
\[ -x = 3 \]
Thus:
\[ x = -3 \]

### Step 3: Find the value of \(y\)
Now substitute \(x = -3\) back into Equation 1 to find \(y\):
\[ 2(-3) + y = 1 \]
This simplifies to:
\[ -6 + y = 1 \]
Adding 6 to both sides:
\[ y = 7 \]

### Step 4: Intersection point
The intersection point of the two lines is \((-3, 7)\).

### Step 5: Find the equation of the line through the origin and the intersection point
We need to find the equation of the line that passes through the points \((0, 0)\) (the origin) and \((-3, 7)\).

Using the two-point form of the equation of a line:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Substituting \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (-3, 7)\):
\[ y - 0 = \frac{7 - 0}{-3 - 0}(x - 0) \]
This simplifies to:
\[ y = \frac{7}{-3}x \]
or
\[ y = -\frac{7}{3}x \]

### Step 6: Rearranging to standard form
To express this in standard form \(Ax + By + C = 0\):
\[ 7x + 3y = 0 \]

### Final Answer
The equation of the straight line is:
\[ 7x + 3y = 0 \]