Rocco May

2023-03-13

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

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

$\frac{1}{a}-\frac{1}{b}=2$

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

$\frac{1}{a}+\frac{1}{b}=2$

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

$\frac{1}{a}-\frac{1}{b}=2$

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

$\frac{1}{a}+\frac{1}{b}=2$

Deon Bowen

Beginner2023-03-14Added 4 answers

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,

$A=\frac{1}{2}\left|\begin{array}{ccc}1& {x}_{1}& {y}_{1}\\ 1& {x}_{2}& {y}_{2}\\ 1& {x}_{3}& {y}_{3}\end{array}\right|$

$A=\frac{1}{2}\left|\begin{array}{ccc}1& a& 0\\ 1& 0& b\\ 1& 1& 1\end{array}\right|$

$\Rightarrow A=\frac{1}{2}\left(a(b-1)+0(1-0)+1(0-b)\right)$

$\Rightarrow A=\frac{1}{2}\left(ab-a-b\right)$

The area of the triangle with vertices $\left(a,0\right),\left(0,b\right)$ and $(1,1)$ is zero as the given points $(a,0)$, $(0,b)$ and $(1,1)$ are collinear.

i.e., $A=0$

$\Rightarrow $$\frac{1}{2}\left(ab-a-b\right)=0$

$\Rightarrow $$ab-a-b=0$

$\Rightarrow $$a+b=ab$

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

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

$A=\frac{1}{2}\left|\begin{array}{ccc}1& {x}_{1}& {y}_{1}\\ 1& {x}_{2}& {y}_{2}\\ 1& {x}_{3}& {y}_{3}\end{array}\right|$

$A=\frac{1}{2}\left|\begin{array}{ccc}1& a& 0\\ 1& 0& b\\ 1& 1& 1\end{array}\right|$

$\Rightarrow A=\frac{1}{2}\left(a(b-1)+0(1-0)+1(0-b)\right)$

$\Rightarrow A=\frac{1}{2}\left(ab-a-b\right)$

The area of the triangle with vertices $\left(a,0\right),\left(0,b\right)$ and $(1,1)$ is zero as the given points $(a,0)$, $(0,b)$ and $(1,1)$ are collinear.

i.e., $A=0$

$\Rightarrow $$\frac{1}{2}\left(ab-a-b\right)=0$

$\Rightarrow $$ab-a-b=0$

$\Rightarrow $$a+b=ab$

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

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