Rocco May

2023-03-13

If the points $\left(a,0\right)$, $\left(0,b\right)$ and $\left(1,1\right)$ are collinear then which of the following is true?
$\frac{1}{a}+\frac{1}{b}=1$
$\frac{1}{a}-\frac{1}{b}=2$
$\frac{1}{a}-\frac{1}{b}=-1$
$\frac{1}{a}+\frac{1}{b}=2$

Deon Bowen

The area $\left(A\right)$ of a triangle with vertices $\left(x1,y1\right),\left(x2,y2\right)$ and $\left(x3,y3\right)$ calculate by the formula,
$A=\frac{1}{2}\left|\begin{array}{ccc}1& {x}_{1}& {y}_{1}\\ 1& {x}_{2}& {y}_{2}\\ 1& {x}_{3}& {y}_{3}\end{array}\right|$
$A=\frac{1}{2}\left|\begin{array}{ccc}1& a& 0\\ 1& 0& b\\ 1& 1& 1\end{array}\right|$
$⇒A=\frac{1}{2}\left(a\left(b-1\right)+0\left(1-0\right)+1\left(0-b\right)\right)$
$⇒A=\frac{1}{2}\left(ab-a-b\right)$
The area of the triangle with vertices $\left(a,0\right),\left(0,b\right)$ and $\left(1,1\right)$ is zero as the given points $\left(a,0\right)$, $\left(0,b\right)$ and $\left(1,1\right)$ are collinear.
i.e., $A=0$
$⇒$$\frac{1}{2}\left(ab-a-b\right)=0$
$⇒$$ab-a-b=0$
$⇒$$a+b=ab$
$⇒$$\frac{a+b}{ab}=\frac{ab}{ab}$
$⇒$$\frac{1}{a}+\frac{1}{b}=1$

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