Question

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

• A. $\sqrt{3}$
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{3}}$

Let us assume side length of equilateral triangle be a.

Using the given information and applying coordinate geometry, we can get coordinates of all points.

$A=(a,0);B=(0,\frac{a}{\sqrt{3}});O=(0,0)$

We can find the coordinates of D either by using the equation OD=DA=a as side lengths of equilateral triangle are all equal; OR we can also use trignometry OR geometry OR various other mathods to find D.

We get $D=(\frac{a}{2},\frac{\sqrt{3}a}{2})$

$\text{We know that Slope between two points}(x_1,y_1)\text{and}(x_2,y_2)\text{is given by}\frac{y_2-y_1}{x_2-x_1}$

$\text{Thus slope between B and D}=\frac{\frac{\sqrt{3}a}{2}-\frac{a}{\sqrt{3}}}{\frac{a}{2}-0}$

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