The perpendicular distance from the point (1, | vedclass

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

Question

(D) The given points $(1, \pi)$,$(1, 0^{\circ})$,and $(1, \frac{\pi}{2})$ are in polar coordinates $(r, 	heta)$.Converting these to Cartesian coordin

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(D) The given points $(1, \pi)$,$(1, 0^{\circ})$,and $(1, \frac{\pi}{2})$ are in polar coordinates $(r, 	heta)$.Converting these to Cartesian coordinates $(x, y) = (r \cos 	heta, r \sin 	heta)$:$(1, \pi) ightarrow (1 \cos \pi, 1 \sin \pi) = (-1, 0)$$(1, 0^{\circ}) ightarrow (1 \cos 0^{\circ}, 1 \sin 0^{\circ}) = (1, 0)$$(1, \frac{\pi}{2}) ightarrow (1 \cos \frac{\pi}{2}, 1 \sin \frac{\pi}{2}) = (0, 1)$The equation of the line passing through $(1, 0)$ and $(0, 1)$ is given by the intercept form $\frac{x}{a} + \frac{y}{b} = 1$,where $a=1$ and $b=1$:$x + y = 1 \Rightarrow x + y - 1 = 0$The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Substituting $(-1, 0)$ into the formula:$d = \frac{|1(-1) + 1(0) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

The perpendicular distance from the point $(1, \pi)$ to the line joining $(1, 0^{\circ})$ and $(1, \frac{\pi}{2})$ (in polar coordinates) is:

Similar Questions

The packing efficiency of the two-dimensional square unit cell shown below is ........... $\%$

What causes a decrease in soil fertility?

In the following diagram,the charge and potential difference across $8\, \mu F$ capacitance will be nearly

Which of the following statements is not correct?

Find the value of $V_{AB}$ in the given circuit.

Vedclass Test Series

Exam Paper Generator

Online Exam Module