A triangle has two fixed vertices A(a, 0) and | vedclass

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(D) The area $s$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $s = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1

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$\frac{a + b}{2}$

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(D) The area $s$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $s = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Given vertices $A(a, 0)$, $B(0, b)$, and $C(x, x)$, we have:$s = \frac{1}{2} |a(b - x) + 0(x - 0) + x(0 - b)|$$s = \frac{1}{2} |ab - ax - bx| = \frac{1}{2} |ab - x(a + b)|$.Since $s$ is the absolute value, $s = \pm \frac{1}{2} (ab - x(a + b))$.Differentiating with respect to $x$:$\frac{ds}{dx} = \pm \frac{1}{2} (-(a + b)) = \mp \frac{a + b}{2}$.Given the options, the magnitude is $\frac{a + b}{2}$.

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

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