Question
Write the perimeter of the triangle formed by the points O(0, 0), A(a, 0), and B(0, b).
✓
Answer
The vertices of a ∆OAB, O(0, 0), A(a, 0), and B(0, b).
Now length of $\text{OA}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(\text{a}-0)^2+(0-0)^2}$
$=\sqrt{\text{a}^2+0^2}=\sqrt{\text{a}^2}=\text{a}$
$\text{OB}=\sqrt{(0-0)^2+(\text{b}-0)^2}$
$=\sqrt{0^2+\text{b}^2}=\sqrt{\text{b}^2}=\text{b}$
$\text{AB}=\sqrt{(0-\text{a})^2+(\text{b}-0)^2}$
$=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore$ Now area of ∆ABC
$=\frac{1}{2}\times\text{OA}\times\text{OB}$
$=\frac{1}{2}\text{ab sq.units}$
Now length of $\text{OA}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(\text{a}-0)^2+(0-0)^2}$
$=\sqrt{\text{a}^2+0^2}=\sqrt{\text{a}^2}=\text{a}$
$\text{OB}=\sqrt{(0-0)^2+(\text{b}-0)^2}$
$=\sqrt{0^2+\text{b}^2}=\sqrt{\text{b}^2}=\text{b}$
$\text{AB}=\sqrt{(0-\text{a})^2+(\text{b}-0)^2}$
$=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore$ Now area of ∆ABC
$=\frac{1}{2}\times\text{OA}\times\text{OB}$
$=\frac{1}{2}\text{ab sq.units}$
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