Find the value of 'a' such that PO = QR where P, Q, and R are the points whose coordinates are (6, -1), (1, 3) and (a, 8) respectively.
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Coordinate of point P,Q and R are P(6,-1),Q(1,3) and R(x,8)
Given, PQ=QR
So by distance formula we have,
Distance between two points =
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
PQ=
(6−1)
2
+(−1−3)
2
=
25+14
=
41
∴PQ
2
=41=QR
2
but, QR
2
=(x−1)
2
+25
41=x
2
+1−2x+25
⇒41=x
2
+1−2x+25
⇒x
2
−2x+26=41
⇒x
2
−2x−15=0
⇒x
2
−5x+3x−15=0
⇒x(x−5)+3(x−5)=0
⇒(x+3)(x−5)=0
⇒x=−3,5
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