Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the equation of a line drawn perpendicular to the line x/4+y/6=1 through the point, where it meets the y-axis.

Q 7: Find the equation of a line drawn perpendicular to the line  \small \frac{x}{4}+\frac{y}{6}=1                           
              through the point, where it meets the \small y-axis. 

Answers (1)

best_answer

The given equation of line is
\small \frac{x}{4}+\frac{y}{6}=1
we can rewrite it as
3x+2y=12
Slope of line 3x+2y=12 , m' = -\frac{3}{2}
Let the Slope of perpendicular line be m
m = -\frac{1}{m'}= \frac{2}{3}
Now, the point of intersection of 3x+2y=12  and x =0  is   (0,6)
equation of line passing through point (0,6) and with slope \frac{2}{3}  is
(y-6)= \frac{2}{3}(x-0)
3(y-6)= 2x
2x-3y+18=0
Therefore, equation of line is   2x-3y+18=0

Posted by

Gautam harsolia

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years
anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question