The joint equation of pair of lines through point `(1, 2)` and perpendicular to the lines given by `2x^(2)-5xy+3y^(2)=0` is

Options

A. `3x^(2)+5xy+2y^(2)-16x+13y-21=0`
B. `3x^(2)+5xy+2y^(2)+16x-13y-21=0`
C. `3x^(2)+5xy+2y^(2)-16x-13y-21=0`
D. `3x^(2)+5xy+2y^(2)-16x-13y+21=0`

Answer

To find the joint equation of the pair of lines through the point (1, 2) and perpendicular to the lines given by the equation \(2x^2 - 5xy + 3y^2 = 0\), we can follow these steps:

### Step 1: Factor the given equation
We start with the equation of the lines:
\[ 2x^2 - 5xy + 3y^2 = 0 \]
We can factor this quadratic equation. To do this, we can rearrange it as follows:
\[ 2x^2 - 3xy - 2xy + 3y^2 = 0 \]
Now, we can group the terms:
\[ x(2x - 3y) - y(2x - 3y) = 0 \]
Factoring out \( (2x - 3y) \):
\[ (2x - 3y)(x - y) = 0 \]
This gives us the two lines:
1. \(y = \frac{2}{3}x\) (from \(2x - 3y = 0\))
2. \(y = x\) (from \(x - y = 0\))

### Step 2: Find the slopes of the lines
The slopes of the lines are:
- For \(y = x\), the slope \(m_1 = 1\).
- For \(y = \frac{2}{3}x\), the slope \(m_2 = \frac{2}{3}\).

### Step 3: Determine the slopes of the perpendicular lines
The slopes of the lines that are perpendicular to these slopes can be calculated using the negative reciprocal:
- The slope \(m_1'\) perpendicular to \(m_1\) is:
\[ m_1' = -\frac{1}{m_1} = -1 \]
- The slope \(m_2'\) perpendicular to \(m_2\) is:
\[ m_2' = -\frac{1}{m_2} = -\frac{3}{2} \]

### Step 4: Write the equations of the lines through the point (1, 2)
Using the point-slope form of the equation of a line, we can write the equations of the lines passing through the point (1, 2).

1. For the slope \(m_1' = -1\):
\[ y - 2 = -1(x - 1) \]
Simplifying this, we get:
\[ y - 2 = -x + 1 \implies x + y - 3 = 0 \]

2. For the slope \(m_2' = -\frac{3}{2}\):
\[ y - 2 = -\frac{3}{2}(x - 1) \]
Simplifying this, we get:
\[ y - 2 = -\frac{3}{2}x + \frac{3}{2} \implies 3x + 2y - 7 = 0 \]

### Step 5: Find the joint equation of the pair of lines
Now, we need to find the joint equation of the two lines:
1. \(x + y - 3 = 0\)
2. \(3x + 2y - 7 = 0\)

To find the joint equation, we multiply the two equations:
\[ (x + y - 3)(3x + 2y - 7) = 0 \]
Expanding this:
\[ 3x^2 + 2xy - 7x + 3xy + 2y^2 - 21y - 9x - 6y + 21 = 0 \]
Combining like terms:
\[ 3x^2 + 5xy + 2y^2 + 6x - 27y + 21 = 0 \]

### Final Answer
The joint equation of the pair of lines through the point (1, 2) and perpendicular to the lines given by \(2x^2 - 5xy + 3y^2 = 0\) is:
\[ 3x^2 + 5xy + 2y^2 + 6x - 27y + 21 = 0 \]