MCQ
If three points $(0, 0), (3,\sqrt{3})$ and $(3,\lambda)$ form an equilateral triangle, then $\lambda=$
- A$2$
- B$-3$
- C$-4$
- ✓None of these.
✓
Answer
Correct option: D.
None of these.
Let the points $(0, 0), (3,\sqrt{3})$ and $(3,\lambda)$ form an equilateral triangle
$AB = BC = CA$
$\Rightarrow A B^2=B C^2=C A^2$
$\text { Now, } A B^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2$
$=(3-0)^2+(\sqrt{3}-0)^2$
$=(3)^2+(\sqrt{3})^2$
$=9+3=12$
$\text{BC}^2=(3-3)^2+(\lambda-\sqrt{3})^2$
$=(0)^2+(\lambda-\sqrt{3})^2=(\lambda-\sqrt{3})^2$
and $\text{CA}^2=(0-3)^2+(0-\lambda)^2$
$=(-3)^2+(-\lambda)^2$
$=9+\lambda^2$
$\text{AB}^2=\text{CA}^2\Rightarrow\ 12=9+\lambda^2$
$\Rightarrow\ \lambda^2=12-9=3$
$\therefore\ \lambda=\pm\sqrt{3}$
$AB = BC = CA$
$\Rightarrow A B^2=B C^2=C A^2$
$\text { Now, } A B^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2$
$=(3-0)^2+(\sqrt{3}-0)^2$
$=(3)^2+(\sqrt{3})^2$
$=9+3=12$
$\text{BC}^2=(3-3)^2+(\lambda-\sqrt{3})^2$
$=(0)^2+(\lambda-\sqrt{3})^2=(\lambda-\sqrt{3})^2$
and $\text{CA}^2=(0-3)^2+(0-\lambda)^2$
$=(-3)^2+(-\lambda)^2$
$=9+\lambda^2$
$\text{AB}^2=\text{CA}^2\Rightarrow\ 12=9+\lambda^2$
$\Rightarrow\ \lambda^2=12-9=3$
$\therefore\ \lambda=\pm\sqrt{3}$
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