If `25p^(2)+9q^(2)-r^(2)-30pq=0`, then a point on the line `px+qy+r=0` is

Options

A. `(-5,3)`
B. `(1,2)`
C. `(0,0)`
D. `(5,3)`

Answer

To solve the equation \( 25p^2 + 9q^2 - r^2 - 30pq = 0 \) and find a point on the line \( px + qy + r = 0 \), we can follow these steps:

### Step 1: Rewrite the Given Equation
We start with the equation:
\[ 25p^2 + 9q^2 - r^2 - 30pq = 0 \]
We can rearrange this equation to isolate \( r^2 \):
\[ r^2 = 25p^2 + 9q^2 - 30pq \]

### Step 2: Factor the Left Side
Next, we can rearrange the terms:
\[ 25p^2 - 30pq + 9q^2 - r^2 = 0 \]
We can group the first three terms:
\[ (5p - 3q)^2 - r^2 = 0 \]
This can be factored using the difference of squares:
\[ (5p - 3q - r)(5p - 3q + r) = 0 \]

### Step 3: Set Each Factor to Zero
From the factored form, we have two equations:
1. \( 5p - 3q - r = 0 \)
2. \( 5p - 3q + r = 0 \)

### Step 4: Solve for Points
From the first equation:
\[ 5p - 3q = r \implies r = 5p - 3q \]
From the second equation:
\[ 5p - 3q = -r \implies r = 3q - 5p \]

### Step 5: Find Points on the Line
We want to find points that satisfy the line equation \( px + qy + r = 0 \).

#### For the first case:
Substituting \( r = 5p - 3q \) into the line equation:
\[ px + qy + (5p - 3q) = 0 \]
This simplifies to:
\[ px + qy + 5p - 3q = 0 \implies px + qy = -5p + 3q \]
Choosing \( x = 1 \) gives:
\[ p(1) + qy = -5p + 3q \implies p + qy = -5p + 3q \implies qy = -6p + 3q \implies y = \frac{-6p + 3q}{q} \]

#### For the second case:
Substituting \( r = 3q - 5p \) into the line equation:
\[ px + qy + (3q - 5p) = 0 \]
This simplifies to:
\[ px + qy + 3q - 5p = 0 \implies px + qy = 5p - 3q \]
Choosing \( x = -1 \) gives:
\[ p(-1) + qy = 5p - 3q \implies -p + qy = 5p - 3q \implies qy = 6p - 3q \implies y = \frac{6p - 3q}{q} \]

### Final Points
From our calculations, we find two points:
1. \( (5, -3) \)
2. \( (-5, 3) \)

### Conclusion
Thus, a point on the line \( px + qy + r = 0 \) is \( (-5, 3) \).