A line is drawn passing through point P(1,2) to cut positive coordinate axes at A and B . Find minimum area of `DeltaPAB`.

Answer

To find the minimum area of triangle PAB formed by a line passing through the point P(1, 2) and cutting the positive coordinate axes at points A and B, we can follow these steps:

### Step 1: Define Points A and B
Let the coordinates of point A (where the line intersects the y-axis) be (0, k) and the coordinates of point B (where the line intersects the x-axis) be (h, 0).

### Step 2: Equation of the Line
Using the intercept form of the equation of a line, we have:
\[ \frac{x}{h} + \frac{y}{k} = 1 \]
Since the line passes through the point P(1, 2), we can substitute these values into the equation:
\[ \frac{1}{h} + \frac{2}{k} = 1 \]

### Step 3: Express k in terms of h
From the equation above, we can express k in terms of h:
\[ \frac{2}{k} = 1 - \frac{1}{h} \]
\[ \frac{2}{k} = \frac{h - 1}{h} \]
Cross-multiplying gives:
\[ 2h = k(h - 1) \]
Thus, we can express k as:
\[ k = \frac{2h}{h - 1} \]

### Step 4: Area of Triangle PAB
The area \( A \) of triangle PAB can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times h \times k \]
Substituting the value of k we found:
\[ A = \frac{1}{2} \times h \times \frac{2h}{h - 1} = \frac{h^2}{h - 1} \]

### Step 5: Find the Minimum Area
To find the minimum area, we need to differentiate A with respect to h and set the derivative to zero:
\[ A(h) = \frac{h^2}{h - 1} \]
Using the quotient rule for differentiation:
\[ A' = \frac{(h - 1)(2h) - h^2(1)}{(h - 1)^2} \]
Simplifying the numerator:
\[ = \frac{2h^2 - 2h - h^2}{(h - 1)^2} = \frac{h^2 - 2h}{(h - 1)^2} \]
Setting the derivative equal to zero to find critical points:
\[ h^2 - 2h = 0 \implies h(h - 2) = 0 \]
Thus, \( h = 0 \) or \( h = 2 \). Since we are interested in positive values, we take \( h = 2 \).

### Step 6: Calculate Minimum Area
Substituting \( h = 2 \) back into the area formula:
\[ A = \frac{2^2}{2 - 1} = \frac{4}{1} = 4 \]

### Conclusion
The minimum area of triangle PAB is \( 4 \) square units.

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