Question

please answer this question​

Answer 1

Step-by-step explanation:

Given:

The three points given below are collinear,

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

To prove:

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

Proof:

Collinear: given set of points lies on a straight line.

Now we are told that the given set of points are collinear hence they will not be able to form a triangle or the area enclosed by all the points will be equal to zero.

The formula of area of traingle in determinant form is

$Area = \frac{1}{2} \begin{vmatrix} x_{1}& y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}$

that can also be written as

$\sf \small Area = \frac{1}{2}  [x_1 (y_2 - y_3) + x_3 (y_1 - y_2) + x_2 (y_3 - y_1)]$

Since the points are collinear L.H.S will be zero,

$\sf \small 0 = \frac{1}{2}  [x_1 (y_2 - y_3) + x_3 (y_1 - y_2) + x_2 (y_3 - y_1)]$

Substituting all the given cordinates in above equation.

$\sf \small 0 = \frac{1}{2}  [a (b - 1) + 1 (0 - b) + 0 (1 - 0)]$

$\sf \small 0 \times 2 = ab - a - b \\ \sf \small 0 = ab - a - b \\ \sf \small a + b = ab \\ \sf \small \: dividing \: both \: sides \: by \: ab \\ \sf \frac{a}{ab} + \frac{b}{ab} = \frac{ab}{ab} \\ \sf \frac{\cancel a}{\cancel ab} + \frac{\cancel b}{a \cancel b} = \cancel \frac{ab}{ab} \\ \sf \frac{1}{a} + \frac{1}{b} = \frac{1}{1}$

$\sf Hence \: proved,$

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

Thanks for joining brainly community!

Answer 2

Use area of triangle formed by using those points is equal to zero

$\tt Area\: of \:triangle =\frac{1}{2}[a(b-1)+0+1(0-b)]$

$\: \: \: \: \: \: \: \: \: \: \:$

$\tt0=ab-a-b$

$\: \: \: \: \: \: \: \: \: \: \:$

$\tt a+b=ab$

Divide both sides by ab

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

$\: \: \: \: \: \: \: \: \: \: \:$

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

Hence proved~