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

The coordinates of the point P on the line 2x+3y+1=0 such that |PAPB| is maximum, where A(2,0) and B(0,2) is

A
(4,3)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(7,5)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(10,7)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(8,5)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B (7,5)
Given Point P(x1,y1) lies on
2x+3y+1=0-----(1)
Given points A(2,0),B(0,2)
maximum value of |PAPB| is equal to the AB
consider PAB is triangle then
from Triangular inequalities
PB+AB>PA
|PAPB|<PA-------(2)
Now PA=(02)2+(20)2
PA=8
PA=22
from eq (2)
22>|PAPB| is similar to
22>eq of line AB
on solving above eq we get
y1=1(x12)
y1=x1+2
x1+y1=2----(3)
Point P lies in line (1)
2x1+3y1=1----(4)
from eq (3) and (4)
x1=7,y1=5
Point P(7,5)

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