Four points `A, B, C, D` lie in that order on the parabola `y = ax^2 +bx+c.` The co-ordinates of `A, B and D` are known as `A(-2, 3); B(-1, 1) and D(2,7).` The co-ordinates of C for which the area of the quadrilateral `ABCD` is greatest is

Options

A. `(1//2,7//4)`
B. `(1//2 ,-7//4)`
C. `(-1//2,7//4)`
D. `(-1//2,-7//4)`

Answer

To solve the problem of finding the coordinates of point C on the parabola \( y = ax^2 + bx + c \) such that the area of quadrilateral ABCD is maximized, we will follow these steps:

### Step 1: Set up the equations based on the given points
We know the coordinates of points A, B, and D:
- \( A(-2, 3) \)
- \( B(-1, 1) \)
- \( D(2, 7) \)

We can substitute these points into the parabola equation \( y = ax^2 + bx + c \) to create a system of equations.

1. For point A:
\[ 3 = a(-2)^2 + b(-2) + c \implies 3 = 4a - 2b + c \quad \text{(Equation 1)} \]

2. For point B:
\[ 1 = a(-1)^2 + b(-1) + c \implies 1 = a - b + c \quad \text{(Equation 2)} \]

3. For point D:
\[ 7 = a(2)^2 + b(2) + c \implies 7 = 4a + 2b + c \quad \text{(Equation 3)} \]

### Step 2: Solve the system of equations
Now we have three equations:
1. \( 4a - 2b + c = 3 \) (1)
2. \( a - b + c = 1 \) (2)
3. \( 4a + 2b + c = 7 \) (3)

We can eliminate \( c \) by subtracting Equation (2) from Equations (1) and (3).

From (1) - (2):
\[ (4a - 2b + c) - (a - b + c) = 3 - 1 \]
\[ 3a - b = 2 \quad \text{(Equation 4)} \]

From (3) - (2):
\[ (4a + 2b + c) - (a - b + c) = 7 - 1 \]
\[ 3a + 3b = 6 \implies a + b = 2 \quad \text{(Equation 5)} \]

### Step 3: Solve for \( a \) and \( b \)
Now we have two new equations:
1. \( 3a - b = 2 \) (4)
2. \( a + b = 2 \) (5)

From Equation (5), we can express \( b \) in terms of \( a \):
\[ b = 2 - a \]

Substituting this into Equation (4):
\[ 3a - (2 - a) = 2 \]
\[ 3a - 2 + a = 2 \]
\[ 4a = 4 \implies a = 1 \]

Now substituting \( a = 1 \) back into Equation (5):
\[ 1 + b = 2 \implies b = 1 \]

### Step 4: Find \( c \)
Now we can find \( c \) by substituting \( a \) and \( b \) back into any of the original equations. Using Equation (2):
\[ 1 - 1 + c = 1 \implies c = 1 \]

Thus, we have:
\[ a = 1, \quad b = 1, \quad c = 1 \]

The equation of the parabola is:
\[ y = x^2 + x + 1 \]

### Step 5: Find coordinates of point C
Let the coordinates of point C be \( C(x, y) \). Since C lies on the parabola:
\[ y = x^2 + x + 1 \]

### Step 6: Calculate the area of quadrilateral ABCD
The area \( A \) of quadrilateral ABCD can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \]
Where \( A(-2, 3), B(-1, 1), C(x, x^2 + x + 1), D(2, 7) \).

Substituting the coordinates into the area formula and simplifying will yield a function of \( x \).

### Step 7: Maximize the area
To find the value of \( x \) that maximizes the area, we differentiate the area function with respect to \( x \) and set the derivative to zero. Solving this will give us the optimal \( x \) coordinate for point C.

After performing the differentiation and solving, we find:
\[ x = \frac{1}{2} \]

### Step 8: Find the corresponding \( y \) coordinate
Substituting \( x = \frac{1}{2} \) back into the parabola equation:
\[ y = \left(\frac{1}{2}\right)^2 + \frac{1}{2} + 1 = \frac{1}{4} + \frac{1}{2} + 1 = \frac{7}{4} \]

### Final Answer
Thus, the coordinates of point C for which the area of quadrilateral ABCD is greatest are:
\[ C\left(\frac{1}{2}, \frac{7}{4}\right) \]