Let `M(2,13/8)` is the circumcentre of `DeltaPQR` whose sides `PQ and PR` are represented by the straight lines `4x-3y = 0 and 4x + y = 16` respectively. The orthocentre of `DeltaPQR` is

Options

A. a. area of `DeltaABC` equals `4`
B. b. area of `DeltaABC` equals `2`
C. c. coordinates of point `A` is `((3)/(2),2)`
D. d. coordinates of point `C` is `((7)/(2), 2)`

Explanation

To find the orthocenter of triangle \( PQR \) given the circumcenter \( M(2, \frac{13}{8}) \) and the equations of the sides \( PQ \) and \( PR \), we can follow these steps:

### Step 1: Identify the equations of the lines
The equations of the lines representing the sides of the triangle are:
1. \( PQ: 4x - 3y = 0 \) (or \( y = \frac{4}{3}x \))
2. \( PR: 4x + y = 16 \) (or \( y = -4x + 16 \))

### Step 2: Find the slopes of the lines
From the equations, we can find the slopes:
- The slope of line \( PQ \) is \( \frac{4}{3} \).
- The slope of line \( PR \) is \( -4 \).

### Step 3: Find the slopes of the altitudes
The altitudes of the triangle are perpendicular to the sides. Therefore, we can find the slopes of the altitudes:
- The slope of the altitude from point \( R \) (perpendicular to \( PQ \)) is the negative reciprocal of the slope of \( PQ \):
\[ m_{altitude\ PQ} = -\frac{3}{4} \]
- The slope of the altitude from point \( Q \) (perpendicular to \( PR \)) is the negative reciprocal of the slope of \( PR \):
\[ m_{altitude\ PR} = \frac{1}{4} \]

### Step 4: Find the equations of the altitudes
Using the circumcenter \( M(2, \frac{13}{8}) \) to find the equations of the altitudes:
1. For the altitude from \( R \) (slope \( -\frac{3}{4} \)):
\[ y - \frac{13}{8} = -\frac{3}{4}(x - 2) \]
Rearranging gives:
\[ y = -\frac{3}{4}x + \frac{3}{2} + \frac{13}{8} \]
Simplifying:
\[ y = -\frac{3}{4}x + \frac{12}{8} + \frac{13}{8} = -\frac{3}{4}x + \frac{25}{8} \]

2. For the altitude from \( Q \) (slope \( \frac{1}{4} \)):
\[ y - \frac{13}{8} = \frac{1}{4}(x - 2) \]
Rearranging gives:
\[ y = \frac{1}{4}x - \frac{1}{2} + \frac{13}{8} \]
Simplifying:
\[ y = \frac{1}{4}x - \frac{4}{8} + \frac{13}{8} = \frac{1}{4}x + \frac{9}{8} \]

### Step 5: Find the intersection of the altitudes
To find the orthocenter, we need to find the intersection of the two altitude lines:
1. Set the equations equal to each other:
\[ -\frac{3}{4}x + \frac{25}{8} = \frac{1}{4}x + \frac{9}{8} \]
2. Multiply through by 8 to eliminate fractions:
\[ -6x + 25 = 2x + 9 \]
3. Solve for \( x \):
\[ -6x - 2x = 9 - 25 \implies -8x = -16 \implies x = 2 \]
4. Substitute \( x = 2 \) back into one of the altitude equations to find \( y \):
\[ y = -\frac{3}{4}(2) + \frac{25}{8} = -\frac{3}{2} + \frac{25}{8} \]
Convert \( -\frac{3}{2} \) to eighths:
\[ -\frac{3}{2} = -\frac{12}{8} \implies y = -\frac{12}{8} + \frac{25}{8} = \frac{13}{8} \]

### Step 6: Conclusion
Thus, the coordinates of the orthocenter \( H \) of triangle \( PQR \) are:
\[ H(2, \frac{13}{8}) \]