The line `x/a+y/b=1` cuts the axes in A and B and a line perpendicular to AB cuts the axes in P and Q. Locus of the points of intersection of AQ and BP is `x^(2)+y^(2)+ax+by=0`.

Answer

To solve the problem, we need to find the locus of the points of intersection of the lines AQ and BP given the line \( \frac{x}{a} + \frac{y}{b} = 1 \). Let's go through the steps systematically.

### Step 1: Identify Points A and B
The line \( \frac{x}{a} + \frac{y}{b} = 1 \) intersects the x-axis and y-axis at points A and B respectively.

- For point A (where \( y = 0 \)):
\[ \frac{x}{a} + \frac{0}{b} = 1 \implies x = a \implies A(a, 0) \]

- For point B (where \( x = 0 \)):
\[ \frac{0}{a} + \frac{y}{b} = 1 \implies y = b \implies B(0, b) \]

### Step 2: Equation of Line Perpendicular to AB
The slope of line AB can be calculated from points A and B. The slope \( m \) is given by:
\[ m = \frac{b - 0}{0 - a} = -\frac{b}{a} \]
The slope of the line perpendicular to AB is the negative reciprocal:
\[ m_{\perp} = \frac{a}{b} \]

Using point-slope form, the equation of the line perpendicular to AB that passes through point A can be written as:
\[ y - 0 = \frac{a}{b}(x - a) \implies y = \frac{a}{b}x - \frac{a^2}{b} \]

### Step 3: Find Points P and Q
To find where this line intersects the axes:

- For point P (where \( y = 0 \)):
\[ 0 = \frac{a}{b}x - \frac{a^2}{b} \implies x = a \implies P(a, 0) \]

- For point Q (where \( x = 0 \)):
\[ y = \frac{a}{b}(0) - \frac{a^2}{b} = -\frac{a^2}{b} \implies Q(0, -\frac{a^2}{b}) \]

### Step 4: Equation of Line BP
Using point B and the slope \( m_{\perp} = \frac{a}{b} \), the equation of line BP is:
\[ y - b = -\frac{b}{a}(x - 0) \implies y = -\frac{b}{a}x + b \]

### Step 5: Find Intersection of AQ and BP
To find the intersection of lines AQ and BP, we need to set their equations equal to each other.

The equation of AQ (using points A and Q):
\[ y = \frac{-\frac{a^2}{b}}{a}(x - a) \implies y = -\frac{a}{b}x + a \]

Setting the equations of AQ and BP equal:
\[ -\frac{a}{b}x + a = -\frac{b}{a}x + b \]

### Step 6: Solve for x and y
Rearranging the equation:
\[ -\frac{a}{b}x + \frac{b}{a}x = b - a \]
Multiplying through by \( ab \):
\[ -a^2x + b^2x = ab(b - a) \]
\[ (b^2 - a^2)x = ab(b - a) \]
\[ x = \frac{ab(b - a)}{b^2 - a^2} \]

Substituting \( x \) back into one of the line equations to find \( y \).

### Step 7: Locus Equation
The locus of the intersection points can be expressed as:
\[ x^2 + y^2 + ax + by = 0 \]

### Summary
The locus of the points of intersection of AQ and BP is given by the equation:
\[ x^2 + y^2 + ax + by = 0 \]