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)

We can solve these equations simultaneously. Let's express \(y\) from Equation 1:
\[ y = 1 - 2x \]

Now, substitute \(y\) in Equation 2:
\[ 3x + 2(1 - 2x) = 5 \]
Expanding this gives:
\[ 3x + 2 - 4x = 5 \]
Combining like terms results in:
\[ -x + 2 = 5 \]
Solving for \(x\):
\[ -x = 5 - 2 \implies -x = 3 \implies x = -3 \]

Now substitute \(x = -3\) back into Equation 1 to find \(y\):
\[ 2(-3) + y = 1 \implies -6 + y = 1 \implies y = 1 + 6 \implies y = 7 \]

Thus, the intersection point is \((-3, 7)\).

### Step 2: Use the point-slope form to find the equation of the line.

We have two points: the intersection point \((-3, 7)\) and the origin \((0, 0)\). We can use the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \(m\) is the slope given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Using the points \((0, 0)\) and \((-3, 7)\):
\[ m = \frac{7 - 0}{-3 - 0} = \frac{7}{-3} = -\frac{7}{3} \]

Now substituting into the point-slope form using point \((0, 0)\):
\[ y - 0 = -\frac{7}{3}(x - 0) \implies y = -\frac{7}{3}x \]

### Step 3: Rearranging the equation.

To express this in standard form:
\[ 7x + 3y = 0 \]

Thus, the equation of the straight line is:
\[ \boxed{7x + 3y = 0} \]