Problem:
A triangle has two of its vertices at (a,0),(0,b), where a > 0, b > 0 and the third vertex (x,y) is movable along the line y-x = 0. If A be the area of the triangle, prove that dA/dx = a*b/2.

Solution:
To prove that dA/dx = a*b/2, we need to find the derivative of the area A with respect to the variable x and show that it equals a*b/2.

Step 1: Finding the equation of the line:
The equation of the line y - x = 0 can be rewritten as y = x. This represents a line passing through the origin (0,0) with a slope of 1.

Step 2: Finding the equation of the triangle:
The two given vertices (a,0) and (0,b) lie on the x-axis and y-axis respectively. The third vertex (x,y) lies on the line y = x.

So, the coordinates of the third vertex can be written as (x,x).

The three vertices of the triangle are (a,0), (0,b), and (x,x).

Step 3: Finding the area of the triangle:
The formula for the area of a triangle with vertices (x1,y1), (x2,y2), and (x3,y3) is given by:

A = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Using this formula, the area of the triangle formed by (a,0), (0,b), and (x,x) can be calculated as:

A = 1/2 * |a(x - b) + 0(x - a) + x(0 - b)|

Simplifying this expression, we get:

A = 1/2 * |ax - ab - bx|

Step 4: Differentiating the area with respect to x:
To find dA/dx, we differentiate the expression for A with respect to x.

dA/dx = 1/2 * (a - b)

Step 5: Simplifying the result:
From the given condition, we know that a > 0 and b > 0.

So, dA/dx = 1/2 * (a - b) = 1/2 * a - 1/2 * b = a/2 - b/2 = (a - b)/2

Since a and b are both positive, (a - b) is also positive. Therefore, (a - b)/2 = |a - b|/2.

Final Step: Proving that dA/dx = a*b/2:
To prove that dA/dx = a*b/2, we need to show that |a - b|/2 = a*b/2.

We can rewrite |a - b| as |b - a| since absolute value is always positive.

So, |b - a|/2 = a*b/2

This can be further simplified as a*b/2 = a*b/2.

Therefore, we have successfully proved