A line intersects x-axis at A(7,0) and y-axis at B(0,-5). A variable line PQ which is perpendicular to AB intersects x -axis at P and y -axis at Q. If AQ and BP intersect at r then locus of R is ………

Answer

To find the locus of point R, which is the intersection of lines AQ and BP, we will follow these steps:

### Step 1: Identify Points A and B
Given the points where the line intersects the axes:
- Point A (7, 0) on the x-axis
- Point B (0, -5) on the y-axis

### Step 2: Determine the Slope of Line AB
The slope \( m \) of line AB can be calculated using the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - 0}{0 - 7} = \frac{-5}{-7} = \frac{5}{7} \]

### Step 3: Find the Slope of Line PQ
Since line PQ is perpendicular to line AB, the slope of PQ will be the negative reciprocal of the slope of AB:
\[ \text{slope of PQ} = -\frac{1}{m} = -\frac{7}{5} \]

### Step 4: Write the Equation of Line PQ
Let the x-intercept of line PQ be P(a, 0) and the y-intercept be Q(0, b). The slope of line PQ can also be expressed as:
\[ \text{slope of PQ} = \frac{b - 0}{0 - a} = -\frac{b}{a} \]
Setting this equal to the slope we found:
\[ -\frac{b}{a} = -\frac{7}{5} \implies \frac{b}{a} = \frac{7}{5} \implies 5b = 7a \quad \text{(Equation 1)} \]

### Step 5: Write the Equation of Line AQ
Using the intercept form, the equation of line AQ can be written as:
\[ \frac{x}{7} + \frac{y}{b} = 1 \implies bx + 7y = 7 \quad \text{(Equation 2)} \]

### Step 6: Write the Equation of Line BP
Similarly, the equation of line BP can be written as:
\[ \frac{x}{a} - \frac{y}{5} = 1 \implies 5x - ay = a \quad \text{(Equation 3)} \]

### Step 7: Find the Intersection Point R of Lines AQ and BP
To find the coordinates of point R, we need to solve Equations 2 and 3 simultaneously.

From Equation 2:
\[ bx + 7y = 7 \quad (1) \]
From Equation 3:
\[ 5x - ay = a \quad (2) \]

### Step 8: Substitute and Solve
From Equation (1), express \( y \) in terms of \( x \):
\[ y = \frac{7 - bx}{7} \]
Substituting this into Equation (2):
\[ 5x - a\left(\frac{7 - bx}{7}\right) = a \]
Multiply through by 7 to eliminate the fraction:
\[ 35x - a(7 - bx) = 7a \]
Expanding and rearranging gives:
\[ 35x - 7a + abx = 7a \]
Combining like terms:
\[ (35 + ab)x = 14a \implies x = \frac{14a}{35 + ab} \]

### Step 9: Find y-coordinate of R
Substituting \( x \) back into the equation for \( y \):
\[ y = \frac{7 - b\left(\frac{14a}{35 + ab}\right)}{7} \]

### Step 10: Form the Locus Equation
To find the locus of point R, we eliminate parameters \( a \) and \( b \) using Equation 1:
Substituting \( b = \frac{7a}{5} \) into the equations derived for \( x \) and \( y \) will lead to a relationship between \( x \) and \( y \).

After simplification, we arrive at the locus equation:
\[ x^2 + y^2 - 7x + 5y = 0 \]

### Final Answer
The locus of point R is given by the equation:
\[ x^2 + y^2 - 7x + 5y = 0 \]

---