(B) The distance $d$ of a point $(x, y)$ from the origin $(0, 0)$ is given by the formula $d = \sqrt{x^2 + y^2}$.Substituting the given coordinates $(

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(B) The distance $d$ of a point $(x, y)$ from the origin $(0, 0)$ is given by the formula $d = \sqrt{x^2 + y^2}$.Substituting the given coordinates $(b \cos \theta, b \sin \theta)$:$d = \sqrt{(b \cos \theta)^2 + (b \sin \theta)^2}$$d = \sqrt{b^2 \cos^2 \theta + b^2 \sin^2 \theta}$$d = \sqrt{b^2(\cos^2 \theta + \sin^2 \theta)}$Since $\cos^2 \theta + \sin^2 \theta = 1$,we get:$d = \sqrt{b^2(1)} = b$.

Class 11 Mathematics Straight Line System of co-ordinates, Distance between two points, Section formulae

Medium

The distance of the point $(b \cos \theta, b \sin \theta)$ from the origin is

A
$b \cot \theta$
B
$b$
C
$b \tan \theta$
D
$b \sqrt{2}$

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Class 11 Questions

Class 11

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Chapter

Straight Line

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System of co-ordinates, Distance between two points, Section formulae

If the line joining the points $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$ is extended to the point $N(x, y)$ such that $AN: NB = b: a$,then

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Find the coordinates of the point which divides the line segment joining the points $(a + b, a - b)$ and $(a - b, a + b)$ externally in the ratio $a : b$.

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What is the distance between the points $(a, 0)$ and $(0, a)$?

Easy

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If the points $A(6, -1)$,$B(1, 3)$ and $C(x, 8)$ are such that $AB = BC$,then $x = $

Easy

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Find the slope of a line,which passes through the origin and the mid-point of the line segment joining the points $P(0, -4)$ and $B(8, 0)$.

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The distance of the point $(b \cos \theta, b \sin \theta)$ from the origin is

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If the line joining the points $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$ is extended to the point $N(x, y)$ such that $AN: NB = b: a$,then

Find the coordinates of the point which divides the line segment joining the points $(a + b, a - b)$ and $(a - b, a + b)$ externally in the ratio $a : b$.

What is the distance between the points $(a, 0)$ and $(0, a)$?

If the points $A(6, -1)$,$B(1, 3)$ and $C(x, 8)$ are such that $AB = BC$,then $x = $

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