The joint equation of pair of lines having slopes 1 and 3 and passing through the origin is

Options

A. `4x^(2)-3x-y^(2)=0`
B. `3x^(2)-4xy+y^(2)=0`
C. `3x^(2)-4xy-y^(2)=0`
D. `3x^(2)=y^(2)`

Explanation

To find the joint equation of the pair of lines having slopes 1 and 3 that pass through the origin, we can follow these steps:

### Step 1: Write the equations of the lines
Since both lines pass through the origin (0, 0), we can use the slope-intercept form of the line equation, which is:

\[ y = mx \]

For the first line with slope \( m_1 = 1 \):
\[ y = 1 \cdot x \]
So, the equation of the first line is:
\[ y = x \]

For the second line with slope \( m_2 = 3 \):
\[ y = 3 \cdot x \]
So, the equation of the second line is:
\[ y = 3x \]

### Step 2: Rewrite the equations in standard form
We can rewrite both equations in the standard form \( Ax + By + C = 0 \).

For the first line \( y = x \):
\[ x - y = 0 \]

For the second line \( y = 3x \):
\[ 3x - y = 0 \]

### Step 3: Find the joint equation
The joint equation of the two lines can be found by multiplying their equations:

1. The first line: \( x - y = 0 \)
2. The second line: \( 3x - y = 0 \)

To find the joint equation, we multiply these two equations:

\[ (x - y)(3x - y) = 0 \]

### Step 4: Expand the product
Now, we expand the product:

\[ x(3x) - x(y) - y(3x) + y(y) = 0 \]

This simplifies to:

\[ 3x^2 - xy - 3xy + y^2 = 0 \]

Combining like terms gives:

\[ 3x^2 - 4xy + y^2 = 0 \]

### Final Result
Thus, the joint equation of the pair of lines is:

\[ 3x^2 - 4xy + y^2 = 0 \]