Class 11MathematicsStraight LineDistance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line
MediumStatement $(A)$: The points $(2, 1)$ and $(-3, 5)$ lie on opposite sides of the line $3x - 2y + 1 = 0$.
Reason $(R)$: The algebraic perpendicular distances from the given points to the line have opposite signs.
Reason $(R)$: The algebraic perpendicular distances from the given points to the line have opposite signs.
- A$A$ and $R$ are both independently true and $R$ is the correct explanation for $A$.
- B$A$ and $R$ are both independently true and $R$ is not the correct explanation for $A$.
- C$A$ is true but $R$ is false.
- D$A$ is false but $R$ is true.
Explore More
Similar Questions
Find the distance between the parallel lines $15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$.
Easy
View SolutionIf a line $l$ passes through $(k, 2k), (3k, 3k)$ and $(3, 1)$,where $k \neq 0$,then the distance from the origin to the line $l$ is
The set of all possible values of $\theta$ in the interval $(0, \pi)$ for which the points $(1, 2)$ and $(\sin \theta, \cos \theta)$ lie on the same side of the line $x+y=1$,is . . . . . . .
If $p$ and $q$ are lengths of the perpendiculars from the origin to the lines $x \sec \theta + y \operatorname{cosec} \theta = k$ and $x \cos \theta - y \sin \theta = k \cos 2\theta$ respectively,then
Reduce the equation $x-y=4$ into the normal form $x \cos \omega + y \sin \omega = p$. Find the perpendicular distance from the origin $(p)$ and the angle between the perpendicular and the positive $x$-axis $(\omega)$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo