pointsFor 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

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to evaluate the area of triangle `Delta ABC` , whose vertices A,B,C are exactly the given points, such that:

`A = (1/2)*|[(a,1,1),(1,2,1),(0,b+1,1)]|`

`A = (1/2)*|2a + b + 1 + 0 - 0 - a(b+1) - 1|`

`A = (1/2)*|a + b - ab|`

The problem provides the information that `1/a + 1/b = 1` , such that:

`1/a + 1/b = 1 => (b + a)/(ab) = 1 => b + a = ab`

Substituting `ab` for `a + b` in equation of area yields:

`A = (1/2)*|ab - ab| => A = (1/2)*0 = 0`

Hence, evaluating the area of triangle `Delta ABC` yields 0, hence, the given points lie on a line.

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 318,958 questions. We can answer yours, too.

Ask a question