Answer :
To find the points of intersection between the line [tex]\( L \)[/tex] and the curve [tex]\( C \)[/tex], we need to solve the system of equations formed by their equations. Here are the given equations:
1. The equation of the line [tex]\( L \)[/tex]:
[tex]\[ x - y = 3 \][/tex]
2. The equation of the curve [tex]\( C \)[/tex]:
[tex]\[ 3x^2 - y^2 + xy = 9 \][/tex]
We will solve these equations simultaneously to find the points where the line and the curve intersect.
### Step 1: Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] using the line equation
From the equation of the line [tex]\( x - y = 3 \)[/tex]:
[tex]\[ y = x - 3 \][/tex]
### Step 2: Substitute [tex]\( y = x - 3 \)[/tex] into the equation of the curve
Substitute [tex]\( y = x - 3 \)[/tex] into [tex]\( 3x^2 - y^2 + xy = 9 \)[/tex]:
[tex]\[ 3x^2 - (x - 3)^2 + x(x - 3) = 9 \][/tex]
### Step 3: Expand and simplify the equation
First, expand [tex]\( (x - 3)^2 \)[/tex]:
[tex]\[ (x - 3)^2 = x^2 - 6x + 9 \][/tex]
Now substitute back:
[tex]\[ 3x^2 - (x^2 - 6x + 9) + x(x - 3) = 9 \][/tex]
Simplify the expression:
[tex]\[ 3x^2 - x^2 + 6x - 9 + x^2 - 3x = 9 \][/tex]
[tex]\[ 3x^2 - x^2 + x^2 + 6x - 3x - 9 = 9 \][/tex]
[tex]\[ 3x^2 + 3x - 9 = 9 \][/tex]
### Step 4: Solve the quadratic equation
Combine like terms:
[tex]\[ 3x^2 + 3x - 18 = 0 \][/tex]
Divide the entire equation by 3:
[tex]\[ x^2 + x - 6 = 0 \][/tex]
To solve this quadratic equation, we can factor it:
[tex]\[ (x + 3)(x - 2) = 0 \][/tex]
This gives the solutions:
[tex]\[ x = -3 \quad \text{or} \quad x = 2 \][/tex]
### Step 5: Find the corresponding [tex]\( y \)[/tex] values
Using the line equation [tex]\( y = x - 3 \)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = -3 - 3 = -6 \][/tex]
Thus, one point of intersection is [tex]\( (-3, -6) \)[/tex].
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2 - 3 = -1 \][/tex]
Thus, another point of intersection is [tex]\( (2, -1) \)[/tex].
### Conclusion
The points of intersection between the line [tex]\( L \)[/tex] and the curve [tex]\( C \)[/tex] are:
[tex]\[ (-3, -6) \quad \text{and} \quad (2, -1) \][/tex]
These are the points [tex]\( P \)[/tex] and [tex]\( Q \)[/tex].
1. The equation of the line [tex]\( L \)[/tex]:
[tex]\[ x - y = 3 \][/tex]
2. The equation of the curve [tex]\( C \)[/tex]:
[tex]\[ 3x^2 - y^2 + xy = 9 \][/tex]
We will solve these equations simultaneously to find the points where the line and the curve intersect.
### Step 1: Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] using the line equation
From the equation of the line [tex]\( x - y = 3 \)[/tex]:
[tex]\[ y = x - 3 \][/tex]
### Step 2: Substitute [tex]\( y = x - 3 \)[/tex] into the equation of the curve
Substitute [tex]\( y = x - 3 \)[/tex] into [tex]\( 3x^2 - y^2 + xy = 9 \)[/tex]:
[tex]\[ 3x^2 - (x - 3)^2 + x(x - 3) = 9 \][/tex]
### Step 3: Expand and simplify the equation
First, expand [tex]\( (x - 3)^2 \)[/tex]:
[tex]\[ (x - 3)^2 = x^2 - 6x + 9 \][/tex]
Now substitute back:
[tex]\[ 3x^2 - (x^2 - 6x + 9) + x(x - 3) = 9 \][/tex]
Simplify the expression:
[tex]\[ 3x^2 - x^2 + 6x - 9 + x^2 - 3x = 9 \][/tex]
[tex]\[ 3x^2 - x^2 + x^2 + 6x - 3x - 9 = 9 \][/tex]
[tex]\[ 3x^2 + 3x - 9 = 9 \][/tex]
### Step 4: Solve the quadratic equation
Combine like terms:
[tex]\[ 3x^2 + 3x - 18 = 0 \][/tex]
Divide the entire equation by 3:
[tex]\[ x^2 + x - 6 = 0 \][/tex]
To solve this quadratic equation, we can factor it:
[tex]\[ (x + 3)(x - 2) = 0 \][/tex]
This gives the solutions:
[tex]\[ x = -3 \quad \text{or} \quad x = 2 \][/tex]
### Step 5: Find the corresponding [tex]\( y \)[/tex] values
Using the line equation [tex]\( y = x - 3 \)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = -3 - 3 = -6 \][/tex]
Thus, one point of intersection is [tex]\( (-3, -6) \)[/tex].
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2 - 3 = -1 \][/tex]
Thus, another point of intersection is [tex]\( (2, -1) \)[/tex].
### Conclusion
The points of intersection between the line [tex]\( L \)[/tex] and the curve [tex]\( C \)[/tex] are:
[tex]\[ (-3, -6) \quad \text{and} \quad (2, -1) \][/tex]
These are the points [tex]\( P \)[/tex] and [tex]\( Q \)[/tex].