Joint equation of two lines through (2,-3) perpendicular to two lines `3x^(2)+xy-2y^(2)=0` is

Options

A. `2x^(2)+xy-3y^(2)-5x-20y-25=0`
B. `-2x^(2)-xy+3y^(2)-5x-20y-25=0`
C. `3x^(2)+xy-2y^(2)-5x-20y-25=0`
D. `2x^(2)+xy-3y^(2)+5x+20y-25=0`

Answer

To find the joint equation of two lines through the point (2, -3) that are perpendicular to the two lines represented by the equation \(3x^2 + xy - 2y^2 = 0\), we can follow these steps:

### Step 1: Factor the given equation
The given equation is:
\[ 3x^2 + xy - 2y^2 = 0 \]
We can factor this equation to find the two lines it represents.

Rearranging gives:
\[ 3x^2 + 3xy - 2xy - 2y^2 = 0 \]
Factoring by grouping:
\[ 3x(x + y) - 2y(x + y) = 0 \]
Factoring out \((x + y)\):
\[ (3x - 2y)(x + y) = 0 \]

### Step 2: Identify the slopes of the lines
From the factored form, we have two lines:
1. \(3x - 2y = 0\) which can be rewritten as \(y = \frac{3}{2}x\) (slope \(m_1 = \frac{3}{2}\))
2. \(x + y = 0\) which can be rewritten as \(y = -x\) (slope \(m_2 = -1\))

### Step 3: Find the slopes of the perpendicular lines
To find the slopes of the lines that are perpendicular to these two lines, we take the negative reciprocal of each slope:
- For \(m_1 = \frac{3}{2}\), the perpendicular slope \(m_1' = -\frac{2}{3}\)
- For \(m_2 = -1\), the perpendicular slope \(m_2' = 1\)

### Step 4: Write the equations of the lines through the point (2, -3)
Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (2, -3)\):

1. For the slope \(m_1' = -\frac{2}{3}\):
\[ y + 3 = -\frac{2}{3}(x - 2) \]
Simplifying:
\[ y + 3 = -\frac{2}{3}x + \frac{4}{3} \]
\[ y = -\frac{2}{3}x + \frac{4}{3} - 3 \]
\[ y = -\frac{2}{3}x - \frac{5}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 3y + 5 + 2x = 0 \quad \Rightarrow \quad 2x + 3y + 5 = 0 \quad \text{(Equation 1)} \]

2. For the slope \(m_2' = 1\):
\[ y + 3 = 1(x - 2) \]
Simplifying:
\[ y + 3 = x - 2 \]
\[ y = x - 5 \]
Rearranging gives:
\[ x - y - 5 = 0 \quad \text{(Equation 2)} \]

### Step 5: Find the joint equation
To find the joint equation of the two lines, we multiply the two equations:
\[ (2x + 3y + 5)(x - y - 5) = 0 \]
Expanding this:
\[ 2x^2 - 2xy - 10x + 3xy - 3y^2 - 15y + 5x - 5y - 25 = 0 \]
Combining like terms:
\[ 2x^2 + xy - 3y^2 - 5x - 20y - 25 = 0 \]

### Final Answer
The joint equation of the two lines is:
\[ 2x^2 + xy - 3y^2 - 5x - 20y - 25 = 0 \]