Let’s consider a triangle ABC drawn between the points (1, 1), (2, 0) and (1, 0). Let the triangle drawn through (0, b) (2, 0) and (0, 0) be DBE. Now the triangle DBE is similar to the triangle ABC by the AA criteria for similarity, as the sides DE and AC are parallel.
So the ratio of the lengths of the corresponding sides is the same for both the triangles.
Now the length of side CB is 1 and that of the side AC is also one. So they are in the ratio (1:1). The ratio of the lengths of the sides EB and the side ED should be (1:1).
The length of the side EB is 2 and the length of the side ED is b.
So 2: b = 1: 1
=> b*1 = 2*1
=> b = 2
Therefore b is equal to 2.
The given line is through A(2,0) , a point of intercept of x axis at 2 and P(1,1) and B (0,b) a point of y axis intercept at y = b.
Draw a perpendicular from A to X axis, which meets xaxis at P' or OP' = Px =1.
Then Triangle AOB and triangle AP'P are similar as , PP' || OB as both are perpendicular to X axis.
Therefore OB/OA = P'P/P'A = 1/(2-1).
OB = (1/1)OA = 1*2 = 2.
Therefore OB = b = 2.