The vertices of the base of an isosceles triangle lie on a parabola `y^2 = 4x` and the base is a part of the line `y = 2x - 4`. If the third vertex of the triangle lies on the x-axis, its coordinates are

Options

A. `((5)/(2), 0)`
B. `(7/2, 0)`
C. `((9)/(2) , 0)`
D. `((11)/(2) , 0)`

Answer

To solve the problem step by step, we will find the coordinates of the third vertex of the isosceles triangle that lies on the x-axis, given that the base vertices lie on a parabola and a line.

### Step 1: Understand the equations
The vertices of the base of the isosceles triangle lie on the parabola given by:
\[ y^2 = 4x \]
And the base is a part of the line given by:
\[ y = 2x - 4 \]

### Step 2: Find the intersection points
To find the points where the line intersects the parabola, we will substitute the equation of the line into the equation of the parabola.

Substituting \( y = 2x - 4 \) into \( y^2 = 4x \):
\[ (2x - 4)^2 = 4x \]

### Step 3: Expand and simplify
Expanding the left side:
\[ 4x^2 - 16x + 16 = 4x \]
Now, rearranging the equation:
\[ 4x^2 - 16x + 16 - 4x = 0 \]
\[ 4x^2 - 20x + 16 = 0 \]

### Step 4: Divide by 4
To simplify, divide the entire equation by 4:
\[ x^2 - 5x + 4 = 0 \]

### Step 5: Factor the quadratic equation
Factoring the quadratic:
\[ (x - 1)(x - 4) = 0 \]
Thus, the solutions for \( x \) are:
\[ x = 1 \quad \text{and} \quad x = 4 \]

### Step 6: Find corresponding \( y \) values
Now, we will substitute these \( x \) values back into the line equation \( y = 2x - 4 \) to find the corresponding \( y \) values.

1. For \( x = 1 \):
\[ y = 2(1) - 4 = 2 - 4 = -2 \quad \Rightarrow \quad (1, -2) \]

2. For \( x = 4 \):
\[ y = 2(4) - 4 = 8 - 4 = 4 \quad \Rightarrow \quad (4, 4) \]

### Step 7: Identify the vertices of the base
The vertices of the base \( B \) and \( C \) of the triangle are:
\[ B(1, -2) \quad \text{and} \quad C(4, 4) \]

### Step 8: Find the coordinates of the third vertex \( A \)
Let the coordinates of the third vertex \( A \) on the x-axis be \( (h, 0) \). Since the triangle is isosceles, the distances \( AB \) and \( AC \) must be equal.

Using the distance formula:
\[ AB = \sqrt{(h - 1)^2 + (0 + 2)^2} = \sqrt{(h - 1)^2 + 4} \]
\[ AC = \sqrt{(h - 4)^2 + (0 - 4)^2} = \sqrt{(h - 4)^2 + 16} \]

Setting \( AB = AC \):
\[ \sqrt{(h - 1)^2 + 4} = \sqrt{(h - 4)^2 + 16} \]

### Step 9: Square both sides
Squaring both sides:
\[ (h - 1)^2 + 4 = (h - 4)^2 + 16 \]

### Step 10: Expand and simplify
Expanding both sides:
\[ h^2 - 2h + 1 + 4 = h^2 - 8h + 16 + 16 \]
\[ h^2 - 2h + 5 = h^2 - 8h + 32 \]
Now, cancel \( h^2 \) from both sides:
\[ -2h + 5 = -8h + 32 \]
Rearranging gives:
\[ 6h = 27 \quad \Rightarrow \quad h = \frac{27}{6} = \frac{9}{2} \]

### Step 11: Final coordinates of vertex \( A \)
Thus, the coordinates of the third vertex \( A \) are:
\[ \left(\frac{9}{2}, 0\right) \]

### Summary
The coordinates of the third vertex of the isosceles triangle are:
\[ \boxed{\left(\frac{9}{2}, 0\right)} \]