two opposite vertices of a square are 3 4 and -1 1 find the coordinates of other vertices
Answers
Let PQRS be a square.
P(3,4) and R(1,−1)
Let co-ordinates of a point Q(x,y)
PQ=QR
Squaring on both sides,
PQ
2
=QR
2
⇒ (x−3)
2
+(y−4)
2
=(x−1)
2
+(y−1)
2
Solving this we get,
⇒ x
2
−3x+9+y
2
−8y+16=x
2
−2x+1+y
2
+2y+1
⇒ 4x+10y=23 ---- ( 1 )
⇒ x=
4
23−10y
----- ( 2 )
Length of the hypotenuse =
2
×side
⇒ PR=
2
PQ
⇒ PR
2
=2PQ
2
⇒ (3−1)
2
+(4+1)
2
=2[(x−3)
2
+(y−4)
2
]
⇒ 4+25=2[x
2
−6x+9+y
2
−8y+16]
⇒ 29=2[x
2
−6x+y
2
−8y+25]
⇒ 29=2x
2
−12x+2y
2
−16y+50
⇒ 29=2(
4
23−10y
)
2
−12(
4
23−10y
)+2y
2
−16y+50 [ From ( 2 ) ]
⇒ 29=
16
2
(529−460y+100y
2
)−69+30y+2y
2
−16y+50
⇒ 29=66.125−57.5y+12.5y
2
−69+30y+2y
2
−16y+50
⇒ 29=47.125−43.5y+14.5y
2
⇒ 14.5y
2
−43.5y+18.125=0
By soving we get,
y=
2
5
and y=
2
1
Substituting value of y in equation ( 1 ) we get,
⇒ x=
2
−1
and x=
2
9
Vertices of a square are (
2
9
,
2
1
) and (
2
−1
,
2
5
)
