Let O=(0,0). Let A and B be points on the X-a | vedclass

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(D) Let $B = (0, a)$. Since $\angle OBA = 60^{\circ}$ and $	riangle OAB$ is a right-angled triangle at $O$,we have $	an(60^{\circ}) = \frac{OA}{OB}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $B = (0, a)$. Since $\angle OBA = 60^{\circ}$ and $	riangle OAB$ is a right-angled triangle at $O$,we have $	an(60^{\circ}) = \frac{OA}{OB} = \frac{OA}{a}$. Thus,$OA = a\sqrt{3}$. So,$A = (a\sqrt{3}, 0)$.Since $	riangle OAD$ is an equilateral triangle with vertices $O(0,0)$ and $A(a\sqrt{3}, 0)$,the third vertex $D$ is located at $(\frac{a\sqrt{3}}{2}, \frac{a\sqrt{3} \cdot \sqrt{3}}{2}) = (\frac{a\sqrt{3}}{2}, \frac{3a}{2})$.The slope of line $DB$ passing through $D(\frac{a\sqrt{3}}{2}, \frac{3a}{2})$ and $B(0, a)$ is given by:$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{3a}{2} - a}{\frac{a\sqrt{3}}{2} - 0} = \frac{\frac{a}{2}}{\frac{a\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$.

List-$I$	List-$II$
$A$. Line passing through $(-4, 3)$ and having intercepts in the ratio $5:3$	$1$. $2x - 5y + 4 = 0$
$B$. Line passing through $P(2, -5)$ such that $P$ bisects the part intercepted between the axes	$2$. $3x + 5y = 3$
$C$. Line parallel to $2x - 3y + 5 = 0$ with $x$-intercept $\frac{2}{5}$ is	$3$. $10x - 15y + 4 = 0$
$D$. Line perpendicular to $5x + 2y + 7 = 0$ with $y$-intercept $\frac{4}{5}$ is	$4$. $10x - 15y = 4$
	$5$. $5x - 2y - 20 = 0$

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

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