Question

A triangle has two fixed vertices A \((a,0)\) and B\((0,b)\) . Let its third vertex C is moving along the line \(x = y\). If \(s\) is the area of triangle ABC, then \(\frac{ds}{dx} =\)

• A. \(a+b\)
• B. \(-(\frac{a+b}{2})\)
• C. \(\frac{a-b}{2}\)
• D. \(\frac{a}{2}\)

Hint:

Write the area using the determinant formula with C = (x, x).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Area of a triangle with vertices \((x_1,y_1), (x_2,y_2), (x_3,y_3)\) is \(\tfrac12\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|\).

Step 2: Set up with C = (x, x):
A = (a, 0), B = (0, b), C = (x, x):
\[ s = \frac12\left[a(b - x) + 0\cdot(x - 0) + x(0 - b)\right] = \frac12\left[ab - (a + b)x\right] \]
This is the area when the vertices are listed in the order that gives a positive determinant.

Step 3: Differentiate with respect to x:
\[ \frac{ds}{dx} = -\frac{a + b}{2} \]

Step 4: Match:
This is option (B). Options (A), (C), (D) have the wrong coefficient or sign for \(x\).

Final Answer:
The area is linear in x with slope -(a + b)/2. \[ \boxed{\text{(B) }-\dfrac{a+b}{2}} \]