Given force field

$\vec F = -x\hat i + y\hat j$

We need the work done in moving the particle from $A(1,0)$ to $B(0,1)$ along the line segment joining them.

Equation of the line segment

The line through $(1,0)$ and $(0,1)$ is

$x+y=1 \quad \Rightarrow \quad y=1-x$

We can parametrize the path using $x$ as parameter:

$\vec r(x)=x\hat i +(1-x)\hat j$

where $x$ goes from $1$ to $0$.

Differential displacement

$d\vec r = dx\hat i + dy\hat j$

Since $y=1-x$,

$dy=-dx$

Hence,

$d\vec r = dx\hat i - dx\hat j$

Force along the path

On the path, $y=1-x$, so

$\vec F = -x\hat i + (1-x)\hat j$

Dot product $\vec F\cdot d\vec r$

$\vec F\cdot d\vec r = (-x)dx + (1-x)(-dx)$

$= -x\,dx -(1-x)dx$

$= -x\,dx -dx + x\,dx = -dx$

Work done

$W=\int_A^B \vec F\cdot d\vec r = \int_{x=1}^{0} (-dx)$

$W = -[x]_{1}^{0} = -(0-1)=1$

Checking options

The work done is

$\boxed{1}$

So the correct option is C.

Comparison with stored answer

Stored correct answer: C

Derived answer: C

They match.