For the three points (a, 1) , (1 ,2) and (0, b +1) if 1/a + 1/b = 1, what can be said about the points?  

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation of the line formed by the points (a, 1) and (0, b +1) is y - 1 = [( b + 1 – 1)/ (0 - a)]* (x – a)

=> y – 1 = (-b/a)( x – a)

=> y – 1 = (-b/a)*x + b

=> y – 1 + (b/a)*x - b = 0

Now, if we substitute y with 2 and x with 1 we get

2 – 1  + (b/a)*1 - b = 0

=> 1 + (b/a) - b = 0

=> (b/b) + (b/a) - b =0

=> 1/b + 1/a - 1 = 0

=> 1/ b + 1/a = 1

Therefore the fact that 1/a + 1/b = 1, implies that (a, 1) , (1 ,2) and (0, b +1) are collinear points.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll calculate the determinant formed by the coordinates of these 3 points:

             |a   1   1|

det A = |1    2   1|

            |0  b+1  1|

det A = a*2*1 + (b+1)*1*1 + 1*1*0 - 1*2*0 - a(b+1) - 1*1*1

det A = 2a + b - ab - a (1)

If the value of this determinant is zero, then the 3 given points are collinear. If it does not, then the 3 points are not located on the same line.

We'll consider the constraint given by enunciation:

1/a + 1/b = 1

We'll multiply by a*b the identity:

b + a = ab (2)

We'll combine like terms and we'll substitute the product a*b from (1) by the sum form (2):

det A = a + b - a - b

We'll eliminate like terms:

det A = 0

Since the determinant is zero, then the 3 points are collinear.

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question