Let A=(1,0) , B=(−1,0),C=(2,0), the locus of a point P such that PB2+PC2=2PA2 is
A
a straight line parallel to x-axis
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
a straight line parallel to y-axis
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
pair of straight line
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
combined equation of coordiante axes
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B a straight line parallel to y-axis Let P=(x,y) Then, PA=√(x−1)2+y2 ...(i) PB=√(x+1)2+y2 ...(ii) PC=√(x−2)2+y2 ...(iii) Hence, PC2+PB2=2PA2 ⇒[(x−2)2+y2]+[(x+1)2+y2]=2(x−1)2+2y2 ⇒(x−2)2+(x+1)2+2y2=2(x−1)2+2y2 ⇒(x−2)2+(x+1)2=2(x−1)2 ⇒(x2−4x+4)+(x2+2x+1)=2(x2−2x+1) ⇒2x2−2x+5=2x2−4x+2 ⇒2x+3=0 ⇒x=−32 Hence, this represents a line parallel to y-axis.