The symmetric form of the equation of the line $x = ay + b$ and $z = cy + d$ is
- A$\frac{x - b}{a} = y = \frac{z - d}{c}$
- B$\frac{x - a}{b} = y = \frac{z - c}{d}$
- C$\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}$
- D$\frac{x - a}{b} = \frac{y}{1} = \frac{z - c}{d}$
Explore More
Similar Questions
If the image of the point $P(1, 0, 3)$ in the line joining the points $A(4, 7, 1)$ and $B(3, 5, 3)$ is $Q(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma$ is equal to
DifficultJEE MAIN 2025
View SolutionIf $2x - y + z = 0 = y - x + 2z = mx - 2y + mz$ represents a line in space,then the value of $m$ is-
The distance of the point $A(7, -2, 11)$ from the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$ measured along the line $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6}$ is:
DifficultJEE MAIN 2024
View SolutionThe perpendicular distance from the point $P(-1, 1, 0)$ to the line joining the points $A(0, 2, 4)$ and $B(3, 0, 1)$ is
The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$,then $14 l^2$ is equal to:
Vedclass 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