Q) If points (a, 0), (0, b), (1,1) are collinear, then 1/a + 1/b is ?
Ans: [I have solved the question by two methods. Please choose the method whichever is in line with your syllabus.]
Method 1:
For points to be collinear, area of triangle formed by the three points is equal to 0.
We know that the area of a triangle formed by the points (x1, y1), (x2, y2), (x3, y3) is given by:
![\frac{1}{2} [x_1(y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)] = 0](https://www.saplingacademy.in/wp-content/ql-cache/quicklatex.com-d954f3f5267d43a565635a3bdb313130_l3.png "Rendered by QuickLaTeX.com")
Hence, the area of the triangle formed by the points (a, 0), (0, b), (1,1) will be:
[a (b – 1) + 0 (1 – 0) + 1 (0 – b)] = 0
∴ [a (b – 1) – b] = 0
∴ a b – a – b = 0
∴ a + b = ab
By dividing ab on both sides, we get:
∴ 
∴ 
Method 2:
Let’s consider that given three points [A (a, 0), B (0, b), C (1,1)] are collinear and point B divides the line AC in ratio of k:1
We know that, by section formula, if a point (x, y) divides the line joining the points (x1, y1) and (x2, y2) in the ratio m : n, then the coordinates of intersection point (x, y) is given by:
, 
Therefore, coordinates of points B, dividing line AC in ratio of k : 1 will be:
B(x,y) = 
B(x,y) = 
Since coordinats of B are given as (0, b)
By comparing x coordinate (or abscissa), we get:
= 0
∴ k + a = 0
∴ k = – a
By comparing y coordinate (or ordinate), we get:
= b
∴ k = b (k + 1)
since k = – a
∴ – a = b (- a + 1)
∴ – a = – a b + b
∴ a + b = ab
By dividing ab on both sides, we get:
∴
∴
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