Let a line `x/a+y/b=1` intersects the x-axis at A and y-axis at B resapectively. A line parallel to it is drawn to intersect the axes in P and Q respectively. The extermities of the lines are joined transversely. If the locus of point of intersection of the line joining them is `x/a=cy/b,` then c is equal to

Answer

To solve the problem step by step, we will analyze the given line and the parallel line, find their intersection points, and then derive the locus of the intersection point.

### Step 1: Identify the given line
The equation of the given line is:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
This line intersects the x-axis and y-axis at points A and B respectively.

### Step 2: Find the coordinates of points A and B
- The x-intercept (point A) occurs when \(y = 0\):
\[ \frac{x}{a} + \frac{0}{b} = 1 \implies x = a \implies A(a, 0) \]
- The y-intercept (point B) occurs when \(x = 0\):
\[ \frac{0}{a} + \frac{y}{b} = 1 \implies y = b \implies B(0, b) \]

### Step 3: Write the equation of the parallel line
A line parallel to the given line can be expressed as:
\[ \frac{x}{a} + \frac{y}{b} = k \]
for some constant \(k\). This line intersects the axes at points P and Q.

### Step 4: Find the coordinates of points P and Q
- The x-intercept (point P) occurs when \(y = 0\):
\[ \frac{x}{a} + \frac{0}{b} = k \implies x = ak \implies P(ak, 0) \]
- The y-intercept (point Q) occurs when \(x = 0\):
\[ \frac{0}{a} + \frac{y}{b} = k \implies y = bk \implies Q(0, bk) \]

### Step 5: Find the equations of lines AQ and BP
- The line AQ connects points A and Q:
- A is at \(A(a, 0)\) and Q is at \(Q(0, bk)\).
- The slope of AQ is:
\[ \text{slope of AQ} = \frac{bk - 0}{0 - a} = -\frac{bk}{a} \]
- The equation of line AQ using point-slope form:
\[ y - 0 = -\frac{bk}{a}(x - a) \implies y = -\frac{bk}{a}x + bk \]

- The line BP connects points B and P:
- B is at \(B(0, b)\) and P is at \(P(ak, 0)\).
- The slope of BP is:
\[ \text{slope of BP} = \frac{0 - b}{ak - 0} = -\frac{b}{ak} \]
- The equation of line BP using point-slope form:
\[ y - b = -\frac{b}{ak}(x - 0) \implies y = -\frac{b}{ak}x + b \]

### Step 6: Find the intersection point of lines AQ and BP
Set the equations of AQ and BP equal to find the intersection point:
\[ -\frac{bk}{a}x + bk = -\frac{b}{ak}x + b \]
Multiply through by \(a\) to eliminate the denominators:
\[ -bkx + abk = -bx + ab \]
Rearranging gives:
\[ bx - bky = ab - abk \]
Factoring out \(b\):
\[ b(x - ky) = ab(1 - k) \]
Thus, we have:
\[ x - ky = a(1 - k) \]

### Step 7: Find the locus of the intersection point
From the equation \(x - ky = a(1 - k)\), we can express it in the form:
\[ \frac{x}{a} = k \cdot \frac{y}{b} + (1 - k) \]
This implies:
\[ \frac{x}{a} = c \cdot \frac{y}{b} \]
where \(c = 1\) when \(k = 1\).

### Conclusion
Thus, the value of \(c\) is:
\[ \boxed{1} \]