On the xy-plane, points P and Q are the two points where the parabola with the equation `y =3x ^(2) + (14)/(3) x - (73)/(3)` and the line with thte equation `y =- 4/3 x -1/3` meet. What is the distance between point P and point Q ?

Options

A. 5
B. 8
C. 10
D. 12

Explanation

To find the distance between points P and Q where the parabola \( y = 3x^2 + \frac{14}{3}x - \frac{73}{3} \) intersects the line \( y = -\frac{4}{3}x - \frac{1}{3} \), we will follow these steps:

### Step 1: Set the equations equal to each other
We start by substituting the equation of the line into the equation of the parabola:
\[ -\frac{4}{3}x - \frac{1}{3} = 3x^2 + \frac{14}{3}x - \frac{73}{3} \]

### Step 2: Rearrange the equation
Next, we rearrange the equation to bring all terms to one side:
\[ 3x^2 + \frac{14}{3}x - \frac{73}{3} + \frac{4}{3}x + \frac{1}{3} = 0 \]
Combine like terms:
\[ 3x^2 + \left(\frac{14}{3} + \frac{4}{3}\right)x - \left(\frac{73}{3} - \frac{1}{3}\right) = 0 \]
This simplifies to:
\[ 3x^2 + \frac{18}{3}x - \frac{72}{3} = 0 \]
Which further simplifies to:
\[ 3x^2 + 6x - 24 = 0 \]

### Step 3: Divide by 3
To simplify the equation, we divide everything by 3:
\[ x^2 + 2x - 8 = 0 \]

### Step 4: Factor the quadratic
Next, we factor the quadratic:
\[ (x + 4)(x - 2) = 0 \]
Setting each factor to zero gives us the solutions:
\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

### Step 5: Find corresponding y-values
Now we substitute these x-values back into the line equation to find the corresponding y-values.

For \( x = 2 \):
\[ y = -\frac{4}{3}(2) - \frac{1}{3} = -\frac{8}{3} - \frac{1}{3} = -\frac{9}{3} = -3 \]
So, point P is \( (2, -3) \).

For \( x = -4 \):
\[ y = -\frac{4}{3}(-4) - \frac{1}{3} = \frac{16}{3} - \frac{1}{3} = \frac{15}{3} = 5 \]
So, point Q is \( (-4, 5) \).

### Step 6: Calculate the distance between points P and Q
Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of points P and Q:
\[ d = \sqrt{((-4) - 2)^2 + (5 - (-3))^2} \]
Calculating the differences:
\[ d = \sqrt{(-6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]

### Final Answer
The distance between points P and Q is \( 10 \).

---