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: Base and height route:
Take AB as the base. $|AB| = \sqrt{a^2+b^2}$. The line AB is $bx + ay - ab = 0$.

Step 2: Height from C:
Distance of $(x,x)$ from AB: $\dfrac{|(a+b)x - ab|}{\sqrt{a^2+b^2}}$. So $s = \tfrac12|(a+b)x - ab|$.

Step 3: Differentiate:
For positions where $(a+b)x - ab$ is negative, $s = \tfrac12(ab - (a+b)x)$ and $\dfrac{ds}{dx} = -\dfrac{a+b}{2}$, option (B).

Final Answer:
ds/dx equals -(a + b)/2. \[ \boxed{\text{(B) }-\dfrac{a+b}{2}} \]