P(x, y), A(3, 4) and B (5, -2) are the vertices of triangle PAB such that |PA|=|PB| and area of ΔPAB=10 sq. units, then PA=k√5 units. Find the value of k.
A
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
√5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 2 Given, |PA|=|PB| ∴PA2=PB2⇒(x−3)2+(y−4)2=(x−5)2+(y+2)2⇒x2−6x+9+y2−8y+16=x2−10x+25+y2+4y+4⇒4x−12y=4⇒x−3y=1…(i) area of ΔPAB=12|x(4+2)+3(−2−y)+5(y−4)|⇒10=12|6x−6−3y+5y−20|⇒20=6x+2y−26⇒6x+2y−26=20or6x+2y−26=−20⇒6x+2y=46or6x+2y=63x+y=23…(2)
or 3x+y=3…(3) Solving (1) and (3), we get x=1,y=0 Solving (1) and (2), we get x=7,y=2 ∴ Coordinates of P are (7, 2) and (1, 0) Length of PA=√(7−3)2+(2−4)2=√16+4=2√5∴2√5=k√5⇒k=2 Also, Length of PA=√(1−3)2+(0−4)2=√4+16=2√5∴2√5=k√5⇒k=2