Step-by-step explanation:
Given:
The three points given below are collinear,
[tex](x_1,y_1) = (a,0)\\(x_2,y_2) = (0,b)\\(x_3,y_3) = (1,1)[/tex]
To prove:
[tex] \frac{1}{a} + \frac{1}{b} = 1[/tex]
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
[tex]Area = \frac{1}{2} \begin{vmatrix} x_{1}& y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}[/tex]
that can also be written as
[tex] \sf \small Area = \frac{1}{2} [x_1 (y_2 - y_3) + x_3 (y_1 - y_2) + x_2 (y_3 - y_1)][/tex]
Since the points are collinear L.H.S will be zero,
[tex]\sf \small 0 = \frac{1}{2} [x_1 (y_2 - y_3) + x_3 (y_1 - y_2) + x_2 (y_3 - y_1)][/tex]
Substituting all the given cordinates in above equation.
[tex]\sf \small 0 = \frac{1}{2} [a (b - 1) + 1 (0 - b) + 0 (1 - 0)] [/tex]
[tex]\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} [/tex]
[tex] \sf Hence \: proved,[/tex]
[tex] \Large {\frac{1}{a} + \frac{1}{b} = 1}[/tex]
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