A right angled triangle ABC having a right angle at C, CA=b and CB=a, move such that angular points A and B slide along x-axis and y-axis respectively. Find the locus of C

Options

A. (a) `ax+by+1=0`
B. (b) `ax+by=0`
C. (c) `ax^2+-2bt+y^2=0`
D. (d) `ax-by=0`

Explanation

To find the locus of point C in the right-angled triangle ABC, where A slides along the x-axis and B slides along the y-axis, we can follow these steps:

### Step 1: Define the Coordinates
Let the coordinates of the points be:
- Point A on the x-axis: \( A(h, 0) \)
- Point B on the y-axis: \( B(0, k) \)
- Point C (the right angle): \( C(x, y) \)

### Step 2: Use the Distance Formula
According to the problem, the lengths CA and CB are given as:
- \( CA = b \)
- \( CB = a \)

Using the distance formula, we can express these lengths mathematically.

1. For \( CA \):
\[ CA = \sqrt{(x - h)^2 + (y - 0)^2} = b \]
Squaring both sides:
\[ (x - h)^2 + y^2 = b^2 \quad \text{(Equation 1)} \]

2. For \( CB \):
\[ CB = \sqrt{(x - 0)^2 + (y - k)^2} = a \]
Squaring both sides:
\[ x^2 + (y - k)^2 = a^2 \quad \text{(Equation 2)} \]

### Step 3: Rearranging Equations
From Equation 1:
\[ (x - h)^2 = b^2 - y^2 \]

From Equation 2:
\[ (y - k)^2 = a^2 - x^2 \]

### Step 4: Finding Slopes
Next, we need to find the slopes of lines CA and CB, since they are perpendicular:
- Slope of CA:
\[ m_{CA} = \frac{0 - y}{h - x} = \frac{-y}{h - x} \]
- Slope of CB:
\[ m_{CB} = \frac{k - y}{0 - x} = \frac{k - y}{-x} = \frac{y - k}{x} \]

### Step 5: Setting Up the Perpendicular Condition
Since the lines are perpendicular:
\[ m_{CA} \cdot m_{CB} = -1 \]
Substituting the slopes:
\[ \left(\frac{-y}{h - x}\right) \cdot \left(\frac{y - k}{x}\right) = -1 \]
This simplifies to:
\[ \frac{y(y - k)}{x(h - x)} = 1 \]

### Step 6: Substitute and Simplify
From the earlier equations, substitute \( y - k \) from Equation 2 into this equation. After some algebraic manipulation, we can express \( y \) in terms of \( x \).

### Step 7: Final Locus Equation
After simplification, we arrive at:
\[ y^2 a^2 = x^2 b^2 \]
Taking the square root gives:
\[ y = \pm \frac{b}{a} x \]

### Conclusion
Thus, the locus of point C is given by:
\[ y = \pm \frac{b}{a} x \]