The $$x$$-coordinate of the incentre of the triangle that has the coordinates of midpoints of its sides as (0, 1), (1, 1) and (1, 0) is

We are told that the three mid-points of the sides of some triangle are $$P(0,1),\;Q(1,1),\;R(1,0).$$ Our first task is to recover the actual vertices of the triangle from these mid-points.

For any triangle, if $$P,Q,R$$ are the mid-points of $$BC,CA,AB$$ respectively, then vectors give the relations

$$A = Q + R - P,\qquad B = P + R - Q,\qquad C = P + Q - R.$$

Substituting the given mid-points:

$$A = (1,1) + (1,0) - (0,1) = (2,0),$$

$$B = (0,1) + (1,0) - (1,1) = (0,0),$$

$$C = (0,1) + (1,1) - (1,0) = (0,2).$$

So the triangle’s vertices are $$A(2,0),\;B(0,0),\;C(0,2).$$

Next we need the lengths of the sides opposite these vertices. Using the distance formula $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},$$ we compute

Side $$a$$ (opposite $$A$$) is $$BC$$: $$a = \sqrt{(0-0)^2+(0-2)^2}=2.$$

Side $$b$$ (opposite $$B$$) is $$CA$$: $$b = \sqrt{(0-2)^2+(2-0)^2}= \sqrt{4+4}=2\sqrt{2}.$$

Side $$c$$ (opposite $$C$$) is $$AB$$: $$c = \sqrt{(2-0)^2+(0-0)^2}=2.$$

For the coordinates of the incentre we use the well-known formula: if the vertices are $$A(x_1,y_1),\;B(x_2,y_2),\;C(x_3,y_3)$$ with opposite sides $$a,b,c,$$ then

$$I\left(\dfrac{ax_1+bx_2+cx_3}{a+b+c},\; \dfrac{ay_1+by_2+cy_3}{a+b+c}\right).$$

We need only the $$x$$-coordinate. Putting in the values we have just found:

Numerator:

$$ax_1 = 2\cdot 2 = 4,\qquad bx_2 = 2\sqrt{2}\cdot 0 = 0,\qquad cx_3 = 2\cdot 0 = 0.$$

Hence $$ax_1+bx_2+cx_3 = 4.$$

Denominator:

$$a+b+c = 2 + 2\sqrt{2} + 2 = 4 + 2\sqrt{2}.$$

Therefore

$$x_I = \dfrac{4}{4 + 2\sqrt{2}} = \dfrac{4}{2(2+\sqrt{2})} = \dfrac{2}{2+\sqrt{2}}.$$

To rationalise the denominator we multiply the numerator and the denominator by $$2-\sqrt{2}:$$

$$x_I = \dfrac{2}{2+\sqrt{2}}\times\dfrac{2-\sqrt{2}}{2-\sqrt{2}} = \dfrac{2(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} = \dfrac{2(2-\sqrt{2})}{4-2} = \dfrac{2(2-\sqrt{2})}{2} = 2-\sqrt{2}.$$

Thus the $$x$$-coordinate of the incentre is $$2-\sqrt{2}.$$

Hence, the correct answer is Option D.