The least distance from the origin to a point | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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,

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

If $p_1$ and $p_2$ are the lengths of the perpendiculars from the origin to the lines $x \sec \alpha + y \csc \alpha = 2a$ and $x \cos \alpha - y \sin \alpha = a \cos 2\alpha$ respectively,then what is the value of $\left( \frac{p_1}{p_2} + \frac{p_2}{p_1} \right)^2$?

If the equation of the straight line passing through the point of intersection of $x+2y-19=0$ and $x-2y-3=0$ and which is at a perpendicular distance of $5$ units from the point $(-2,4)$ is $5x+by+c=0$,then a possible value of $5+b+c$ is

The number of integer values of $m$,for which the $x$-coordinate of the point of intersection of the lines $3x + 4y = 9$ and $y = mx + 1$ is also an integer,is

The point on the line $3x + 4y = 5$ which is equidistant from $(1, 2)$ and $(3, 4)$ is

If $l, m, n$ are in arithmetic progression,then the straight line $lx + my + n = 0$ will always pass through the point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module