Question

If the point (a, ), (0, b) and (1, 1) are collinear then which of the following is true?

Answer 1

Hint: First take all the three points and use the formula of area of triangle to find its area and equate it to ‘0’ as we know that area of line is 0.

Complete step-by-step answer:

In the question we are given three points (a, b), (0, b) and (1, 1) which are said to be collinear.
Now as know that the area of line is always ‘0’ we can use this fact by taking or finding the area of the three points using the area formula which is:
$\dfrac{1}{2}\left( {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right)$
As they are in a line the area will be zero if the point given are in form of $\left({{x}_{1}}, {{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$. So we will use points (a, 0), (0, b) and (1, 1) to find the relation between a and b.
Instead of $\left({{x}_{1}}, {{y}_{1}} \right)$ we will substitute (a, 0), $\left( {{x}_{2}},{{y}_{2}} \right)$ we will substitute (0, b) and $\left( {{x}_{3,}}{{y}_{3}} \right)$ we will substitute (1, 1).
So the area is,
$\dfrac{1}{2}\left( a\left( b-1 \right)+0\left( 1-0 \right)+1\left( 0-b \right) \right)=0$
Now on cross multiplication we get,
a(b – 1) + 0+(-b) = 0
On further simplification we get,
ab - a – b = 0
Now taking the terms a, b from left to right hand side we get,
ab = a + b.
Now by dividing by ‘ab’ throughout the equation we get,
$1=\dfrac{1}{b}+\dfrac{1}{a}$ .
So by rearranging and reversing the equation we get,
$\dfrac{1}{a}+\dfrac{1}{b}=1$

Note: We can do this problem by another method by using the fact they are collinear, if we take (a, 0), (0, b) and (0, b) (1, 1) at a time and calculate their slopes and compare it we will get the relation between a, b.