The least distance from the origin to a point | vedclass

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(D) Let the point be $(x, y)$ on the line $y=x+3$.Since the slope of the line is $m=1$,the angle of inclination is $	heta=45^{\circ}$.The point $(x,

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$\sqrt{13-6 \sqrt{2}}$

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(D) Let the point be $(x, y)$ on the line $y=x+3$.Since the slope of the line is $m=1$,the angle of inclination is $	heta=45^{\circ}$.The point $(x, y)$ lies on the line $y=x+3$ and is at a distance of $2$ units from $(0,3)$.Using the parametric form of a line passing through $(0,3)$ with angle $	heta=45^{\circ}$:$\frac{x-0}{\cos 45^{\circ}} = \frac{y-3}{\sin 45^{\circ}} = r$,where $r = \pm 2$.This gives $x = r \cos 45^{\circ} = \pm 2 	imes \frac{1}{\sqrt{2}} = \pm \sqrt{2}$ and $y = 3 + r \sin 45^{\circ} = 3 \pm \sqrt{2}$.The two possible points are $P_1 = (\sqrt{2}, 3+\sqrt{2})$ and $P_2 = (-\sqrt{2}, 3-\sqrt{2})$.The distance $D$ from the origin $(0,0)$ is given by $D^2 = x^2 + y^2$.For $P_1$: $D_1^2 = (\sqrt{2})^2 + (3+\sqrt{2})^2 = 2 + (9 + 2 + 6\sqrt{2}) = 13 + 6\sqrt{2}$.For $P_2$: $D_2^2 = (-\sqrt{2})^2 + (3-\sqrt{2})^2 = 2 + (9 + 2 - 6\sqrt{2}) = 13 - 6\sqrt{2}$.The least distance is $\sqrt{13-6\sqrt{2}}$.

The least distance from the origin to a point on the line $y=x+3$ which lies at a distance of $2$ units from $(0,3)$ is

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