The distance between A(cos theta, 0) and B(0, | vedclass

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

Question

(B) The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Given p

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Given points are $A(\cos 	heta, 0)$ and $B(0, \sin 	heta)$.Substituting the coordinates into the formula:$d = \sqrt{(0 - \cos 	heta)^2 + (\sin 	heta - 0)^2}$$d = \sqrt{(-\cos 	heta)^2 + (\sin 	heta)^2}$$d = \sqrt{\cos^2 	heta + \sin^2 	heta}$Using the trigonometric identity $\sin^2 	heta + \cos^2 	heta = 1$,we get:$d = \sqrt{1} = 1$.

The distance between $A(\cos \theta, 0)$ and $B(0, \sin \theta)$ is ...........

Similar Questions

Find the length of the median through the vertex $(-1, 3)$ of the triangle having vertices $(-1, 3)$,$(1, -1)$,and $(5, 1)$.

If $P(at^{2}, 2at)$,$Q(\frac{a}{t^{2}}, \frac{-2a}{t})$ and $S(a, 0)$ are three points,show that $\frac{1}{SP} + \frac{1}{SQ}$ is independent of $t$.

Prove that $A (1, 7)$,$B (2, 4)$,and $C (5, 5)$ are the vertices of an isosceles right triangle.

The perpendicular distance of the $Y$-axis from the point $(-2, 5)$ is.......

The point which lies on the perpendicular bisector of the line segment joining the points $A (-2,-5)$ and $B (2,5)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module