If the points P (12,8), Q (-2, a) and R (6,0) are the vertices of a right angled triangle PQR, where `angleR = 90^(@)`, the value of a is 

Options

A. 6
B. `-2`
C. `-4`
D. `-6`

Answer

To find the value of \( a \) such that the points \( P(12,8) \), \( Q(-2,a) \), and \( R(6,0) \) form a right-angled triangle at point \( R \), we can follow these steps:

### Step 1: Calculate the slope of line segment \( PR \)
The slope of a line segment between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]
For points \( P(12,8) \) and \( R(6,0) \):
- \( x_1 = 12, y_1 = 8 \)
- \( x_2 = 6, y_2 = 0 \)

Calculating the slope of \( PR \):
\[ \text{slope of } PR = \frac{0 - 8}{6 - 12} = \frac{-8}{-6} = \frac{4}{3} \]

### Step 2: Calculate the slope of line segment \( QR \)
For points \( Q(-2,a) \) and \( R(6,0) \):
- \( x_1 = -2, y_1 = a \)
- \( x_2 = 6, y_2 = 0 \)

Calculating the slope of \( QR \):
\[ \text{slope of } QR = \frac{0 - a}{6 - (-2)} = \frac{-a}{6 + 2} = \frac{-a}{8} \]

### Step 3: Set up the equation for perpendicular slopes
Since \( PR \) and \( QR \) are perpendicular, the product of their slopes should equal -1:
\[ \text{slope of } PR \times \text{slope of } QR = -1 \]
Substituting the slopes we calculated:
\[ \frac{4}{3} \times \frac{-a}{8} = -1 \]

### Step 4: Solve for \( a \)
Now, we can solve the equation:
\[ \frac{4 \cdot -a}{3 \cdot 8} = -1 \]
This simplifies to:
\[ \frac{-4a}{24} = -1 \]
Multiplying both sides by -24:
\[ 4a = 24 \]
Dividing both sides by 4:
\[ a = 6 \]

### Conclusion
The value of \( a \) is \( 6 \).

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