To find the area of the triangle bounded by the line \( x = 4 \) and the lines defined by the equation \( y^2 - x^2 + 2x = 1 \), we will follow these steps:
### Step 1: Rearranging the given equation
The equation given is:
\[
y^2 - x^2 + 2x = 1
\]
Rearranging this, we get:
\[
y^2 = x^2 - 2x + 1
\]
This can be factored as:
\[
y^2 = (x - 1)^2
\]
### Step 2: Finding the equations of the lines
Taking the square root of both sides, we find:
\[
y = x - 1 \quad \text{and} \quad y = -(x - 1) = 1 - x
\]
Thus, we have two lines:
1. \( y = x - 1 \)
2. \( y = 1 - x \)
### Step 3: Finding the intersection point of the lines
To find the intersection of these two lines, we set them equal to each other:
\[
x - 1 = 1 - x
\]
Solving for \( x \):
\[
2x = 2 \implies x = 1
\]
Substituting \( x = 1 \) back into either equation to find \( y \):
\[
y = 1 - 1 = 0
\]
Thus, the intersection point \( P \) is \( (1, 0) \).
### Step 4: Finding points of intersection with \( x = 4 \)
Next, we find the points where these lines intersect the line \( x = 4 \).
For the line \( y = x - 1 \):
\[
y = 4 - 1 = 3
\]
Thus, the point \( Q \) is \( (4, 3) \).
For the line \( y = 1 - x \):
\[
y = 1 - 4 = -3
\]
Thus, the point \( R \) is \( (4, -3) \).
### Step 5: Finding the area of triangle \( PQR \)
Now we have the vertices of the triangle:
- \( P(1, 0) \)
- \( Q(4, 3) \)
- \( R(4, -3) \)
To find the area of triangle \( PQR \), we can use the formula for the area of a triangle given by vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
\text{Area} = \frac{1}{2} \left| 1(3 - (-3)) + 4((-3) - 0) + 4(0 - 3) \right|
\]
Calculating each term:
\[
= \frac{1}{2} \left| 1(6) + 4(-3) + 4(-3) \right|
\]
\[
= \frac{1}{2} \left| 6 - 12 - 12 \right|
\]
\[
= \frac{1}{2} \left| 6 - 24 \right| = \frac{1}{2} \left| -18 \right| = \frac{1}{2} \times 18 = 9
\]
Thus, the area of the triangle \( PQR \) is \( 9 \) square units.
### Final Answer
The area of the triangle is \( \boxed{9} \) square units.
