If R (x, y) is a point on the line segment joining the points

The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula,

$\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

It is said that the point R(x, y) lies on the line segment joining the points P(a, b) and Q(b, a). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.

$\Delta=\frac{1}{2}\left|\left(a y+x a+b^{2}\right)-\left(x b+b y+a^{2}\right)\right|$

$0=\frac{1}{9}\left|a y+x a+b^{2}-x b-b y-a^{2}\right|$

$0=a y+x a+b^{2}-x b-b y-a^{2}$

$a^{2}-b^{2}=a x+a y-b x-b y$

$(a+b)(a-b)=(a-b)(x+y)$

$(a+b)=(x+y)$

Hence under the given conditions we have proved that $x+y=a+b$.