Question

If the points (a,0),(0,b) and (1,1) are collinear ,show that (1/a)+(1/b)=1

Answer 1

use area of triangle form by these three points is equal to zero

area of t triangle=1/2 {a (b-1)+0+1 (0-b)}
0=ab-a-b
a+b=ab
divide both side by ab

a/ab+b/ab=ab/ab

1/b+1/a=1
hence
1/a+1/b=1

Answer 2

Answer:

Here given three points are (a,0),(0,b) and (1,1) are collinears.

We know,

if $(x_1,y_1),(x_2,y_2),(x_3,y_3)$

are three points and they are collinear then satisfy following conditions

$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0......(1)$

Here,

$(x_1,y_1) = (a,0) \\ (x_2,y_2) = (0,b) \\ (x_3,y_3) = (1,1)$

From equation (1),

$a(b - 1) + 0(1 - 0) + 1(0 - b) = 0 \\ a(b - 1) + 0 + 1(0 - b) = 0 \\ ab - a + 0 - b = 0 \\ ab - a - b = 0 \\ ab = a + b$

We are dividing both side by ab,

$\frac{a}{ab} + \frac{b}{ab} = 1 \\ \frac{1}{b} + \frac{1}{a} = 1$

Hence proved.

Some important mathematics formulas :

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Two more problems about Collinear:

https://brainly.in/question/7031149

https://brainly.in/question/10870951