# If the points (x , 1) , (1 , 2) and (0 , y +1) are collinear, show that 1/x + 1/y = 1.

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### 2 Answers

If the given point are collinear then the mathematical determinant formed with these coordinates has to be equal to zero.

This determinant will have on the first column, the coordinates of x, of collinear given points, and on the second column, the coordinates for y, for the three collinear points. And on the third column, it will have the value 1.

Det A, with 3 lines and 3 columns will,be:

x 1 1

1 2 1

0 y+1 1

**Collinear condition for the 3 points, is that the Det A= 0!**

We'll calculate Det A=2*x*1 + (y+1)*1*1 + 1*1*0 - 1*2*0 - x*(y+1) - 1*1*1

Det A=x+y-x*y

But from the Collinear condition, Det A=0.

x+y-x*y=0

**x+y=x*y - Collinear condition**

Now we'll try to prove that 1/x + 1/y=1

In order to sum the 2 ratio, they will have to have the same denominator, which it will be x*y. For this reason,we'll amplify the first ratio with y and the second, with x, and to the right of equal sign, we'll amplify the value 1 with x*y. The result will be

x+y=xy

Given the collinear condition, which it was written above, the relation was demonstrated!

Since the points, (*x* , 1) , (1 , 2) and (0 , *y *+1) are said to be collinear, the points (0,y+1) should lie on the line joining (x,1) and (1,2).

The line joing the points (x,1) and (1,2) is given by:

Y-1 =[ (2-1)/(1-x)] {X-x}.................(1). Since this line passes through the point (0, y+1) , the coordinates (0, y+1) should satisfy eq (1). So,

y+1-1 = 1/(1-x) (0-x). Or

y(1-x) = -x Or

-xy =-x-y. Dividing by -xy,

1 = 1/x + 1/y

Note: It is custumory to write variable (or running ) coordinates as x and y and fixed coordintes of a particular point as (x1,y1). Students, particularly the beginners of analytical coordinate geometry, may get confusion if they deviate from the standard notations and customs followed in texts of maths . In this example , therefore, we have chosen X and Y as variable coordinates and the point (x, 1) and (0,y+1) etc as fixed points.