CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The perpendicular distance from the point of intersection of the lines 3x+2y+4=0, 2x+5y1=0 to the line 7x+24y15=0 is

A
23
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
17
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
15
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
25
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 15

Point of intersection of 3x+2y+4=0 & 2x+5y1=0 is

6x+4y+8=0

6x15y+3=0

11y+11=0

y=1 and x=2

The point of intersection is (2,1).

So, distance of (7x+24y15=0) line from (2,1) is

d=(7×2+241525)

=(101525)

=525

d=15.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Line and a Point
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon