MCQ
If points $(a, 0), (0, b)$ and $(1, 1)$ are collinear, then $\frac{1}{\text{a}}+\frac{1}{\text{b}}=$
- ✓$1$
- B$2$
- C$0$
- D$-1$
✓
Answer
Correct option: A.
$1$
The area of triangle whose vertices are $(a, 0), (0, b)$ and $(1, 1).$
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[\text{a}(\text{b}-1)+0(1-0)+1(0-\text{b})]$
$=\frac{1}{2}[\text{ab}-\text{a}+0-\text{b}]$
$=\frac{1}{2}(\text{ab}-\text{a}-\text{b})$
$\because$ The points are collinear
$\therefore\ \frac{1}{2}(\text{ab}-\text{a}-\text{b})=0$
$\Rightarrow\ \text{ab}-\text{a}-\text{b}=0$
$\Rightarrow\ \text{ab}=\text{a}+\text{b}$
$\Rightarrow\ \frac{\text{a}+\text{b}}{\text{ab}}=1$
$\Rightarrow\ \frac{\text{a}}{\text{ab}}+\frac{\text{b}}{\text{ab}}=1$
$\Rightarrow\ \frac{1}{\text{b}}+\frac{1}{\text{a}}=1$
$\Rightarrow\ \frac{1}{\text{a}}+\frac{1}{\text{b}}=1$
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[\text{a}(\text{b}-1)+0(1-0)+1(0-\text{b})]$
$=\frac{1}{2}[\text{ab}-\text{a}+0-\text{b}]$
$=\frac{1}{2}(\text{ab}-\text{a}-\text{b})$
$\because$ The points are collinear
$\therefore\ \frac{1}{2}(\text{ab}-\text{a}-\text{b})=0$
$\Rightarrow\ \text{ab}-\text{a}-\text{b}=0$
$\Rightarrow\ \text{ab}=\text{a}+\text{b}$
$\Rightarrow\ \frac{\text{a}+\text{b}}{\text{ab}}=1$
$\Rightarrow\ \frac{\text{a}}{\text{ab}}+\frac{\text{b}}{\text{ab}}=1$
$\Rightarrow\ \frac{1}{\text{b}}+\frac{1}{\text{a}}=1$
$\Rightarrow\ \frac{1}{\text{a}}+\frac{1}{\text{b}}=1$
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