CBSE BoardEnglish MediumSTD 10MathsCo-ordinate Geometry3 Marks
Question
Prove that the points (a, 0), (0, b) and (1, 1) are collinear if $\frac{1}{\text{a}}+\frac{1}{\text{b}}=1.$
✓
Answer
Let the points are A(a, 0), B(0, b) and C(1, 1) which form a triangle.
$\therefore$ Area of $\triangle\text{ABC}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[\text{a}(\text{b}-1)+0(1-0)+1(0-\text{b})]$
$=\frac{1}{2}[\text{ab}-\text{a}+0+1(-\text{b})]=\frac{1}{2}[\text{ab}-\text{a}-\text{b}]$
If the points are collinear, then
area of $\triangle\text{ABC}=0$
$\Rightarrow\ \frac{1}{2}(\text{ab}-\text{a}-\text{b})=0$
$\Rightarrow\ \text{ab}-\text{a}-\text{b}=0$
$\Rightarrow\ \text{ab}=\text{a}+\text{b}$
Dividing by ab,
$\frac{\text{ab}}{\text{ab}}=\frac{\text{a}}{\text{ab}}+\frac{\text{b}}{\text{ab}}$
$\Rightarrow\ 1=\frac{1}{\text{b}}+\frac{1}{\text{a}}\Rightarrow\ \frac{1}{\text{a}}+\frac{1}{\text{b}}=1$
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