Let a and b be two non -zero reals such that `aneb`. Then the equation of the line passing through origin and point of intersection of `(x)/(a)+(y)/(b)=1 and (x)/(b)+(y)/(a)=1` is

Options

A. `ax+by=0`
B. `bx+ay=0`
C. `y-x=0`
D. `x+y=0`

Answer

To solve the problem, we need to find the equation of the line passing through the origin and the point of intersection of the two given lines:

1. **Identify the equations of the lines:**
The equations given are:
\[ \frac{x}{a} + \frac{y}{b} = 1 \quad \text{(1)} \]
\[ \frac{x}{b} + \frac{y}{a} = 1 \quad \text{(2)} \]

2. **Convert the equations into standard form:**
Rearranging equation (1):
\[ bx + ay = ab \quad \text{(3)} \]
Rearranging equation (2):
\[ ax + by = ab \quad \text{(4)} \]

3. **Multiply the equations to eliminate one variable:**
Multiply equation (3) by \(a\) and equation (4) by \(b\):
\[ a(bx + ay) = a(ab) \implies abx + a^2y = a^2b \quad \text{(5)} \]
\[ b(ax + by) = b(ab) \implies abx + b^2y = ab^2 \quad \text{(6)} \]

4. **Subtract the two equations:**
Subtract equation (5) from equation (6):
\[ (abx + b^2y) - (abx + a^2y) = ab^2 - a^2b \]
This simplifies to:
\[ (b^2 - a^2)y = ab(b - a) \]

5. **Solve for \(y\):**
Assuming \(b \neq a\), we can divide both sides by \(b - a\):
\[ y = \frac{ab}{a + b} \]

6. **Substitute \(y\) back to find \(x\):**
Substitute \(y\) into equation (3) to find \(x\):
\[ bx + a\left(\frac{ab}{a + b}\right) = ab \]
Simplifying gives:
\[ bx + \frac{a^2b}{a + b} = ab \]
Rearranging:
\[ bx = ab - \frac{a^2b}{a + b} \]
\[ bx = \frac{ab(a + b) - a^2b}{a + b} = \frac{ab^2}{a + b} \]
Thus,
\[ x = \frac{ab}{a + b} \]

7. **Point of intersection:**
The point of intersection is:
\[ \left(\frac{ab}{a + b}, \frac{ab}{a + b}\right) \]

8. **Equation of the line through the origin and the intersection point:**
The slope \(m\) of the line through the origin \((0, 0)\) and the point \(\left(\frac{ab}{a + b}, \frac{ab}{a + b}\right)\) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{ab}{a + b} - 0}{\frac{ab}{a + b} - 0} = 1 \]

9. **Equation of the line:**
Using the point-slope form of the line:
\[ y - 0 = 1(x - 0) \implies y = x \]

Thus, the equation of the line passing through the origin and the point of intersection of the two lines is:
\[ y - x = 0 \]