Let O=(0,0) : let A and B be points respectively on X -axis and Y -axis such that ∠ O B A=60^{circ

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9.Straight Line

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Let $O=(0,0)$ : let $A$ and $B$ be points respectively on $X$-axis and $Y$-axis such that $\angle O B A=60^{\circ}$. Let $D$ be a point in the first quadrant such that $A D$ is an equilateral triangle. Then, the slope of $D B$ is

A

$\sqrt{3}$

B

$\sqrt{2}$

C

$\frac{1}{\sqrt{2}}$

D

$\frac{1}{\sqrt{3}}$

(KVPY-2016)

Solution

(d)

$\angle O B A=60^{\circ}$

$\angle O A B=30^{\circ}$

$O A D$ is an equilateral triangle.

$\therefore \quad \angle A O D=\angle O D A=\angle O A D=60^{\circ}$

$O A=A D=O D$

Let $B(0, e )$

$\therefore \quad A(\sqrt{3} a, 0)$

D $\left(\frac{\sqrt{3}}{2} a, \frac{3}{2} a\right)$

Slope of $B D=\frac{\frac{3}{2}-a}{\frac{\sqrt{3}}{2} a}=\frac{1}{\sqrt{3}}$

Standard 11

Mathematics

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