19. If two vertices of a square are (5, 4) and (1-6), then find the coordinates
of its remaining two vertices.
20. The vertices of a AABC are A(5.5) B(15) and C19.1). A line is drawn to
Answers
Answer:
ANSWER
Solution−
ABCDisasquare.
SoAB=BC=CD=DA&thediagonalsAC=BD..........(i).
Weshallapplydistanceformulatogettheabovelengths.
d=
(x
1
−x
2
)
2
+(y
1
−y
2
)
2
.
NowΔABCisarightonewithACashypotenuse(∠ABC=90
o
).
∴AC=
(5−1)
2
+(4+6)
2
units=
116
units=BD(byi)........(ii).
Weknowthatsideofasquare=
2
diagonal
.
∴AB=BC=CD=DA=
2
116
units=
58
units.........(iii).
Now,usingthedistanceformulad=
(x
1
−x
2
)
2
+(y
1
−y
2
)
2
,
AB
2
=BC
2
⟹(x
1
−5)
2
+(y
1
−4)
2
=(x
1
−1)
2
+(y
1
+6)
2
⟹8x
1
+20y
1
−4=0
⟹2x
1
+5y
1
−1=0
⟹y=
5
1−2x
1
........(iv).
∴AB
2
+BC
2
=AC
2
⟹(x
1
−5)
2
+(y
1
−4)
2
+58=116(fromii&iii)
⟹(x
1
−5)
2
+(y
1
−4)
2
=58
⟹(x
1
−5)
2
+(
5
1−2x
1
−4)
2
=58
⟹29x
1
2
−174x
1
−464=0
⟹x
1
2
−6x
1
−16=0
⟹(x
1
−8)(x
1
+2)=0
⟹x
1
=(8,−2).
So,from(iv),
y
1
=(
5
1−2×8
,
5
1−2(−2)
)=(−3,1).
∴B(x
1
,y
1
)=(8,−3)and(−2,1).
SimilarlyAD
2
=DC
2
⟹(x
2
−5)
2
+(y
2
−4)
2
=(x
2
−1)
2
+(y
2
+6)
2
⟹8x
2
+20y
2
−4=0
⟹2x
2
+5y
2
−1=0
⟹y
2
=
5
1−2x
2
........(iv).
∴AD
2
+DC
2
=AC
2
⟹(x
2
−5)
2
+(y
2
−4)
2
+58=116(fromii&iii)
⟹(x
2
−5)
2
+(y
2
−4)
2
=58
⟹(x
2
−5)
2
+(
5
1−2x
2
−4)
2
=58
⟹29x
2
2
−174x
2
−464=0
⟹x
2
2
−6x
2
−16=0
⟹(x
2
−8)(x
2
+2)=0
⟹x
2
=(8,−2).
So,from(iv),
y
2
=(
5
1−2×8
,
5
1−2(−2)
)=(−3,1).
∴D(x
2
,y
2
)=(8,−3)and(−2,1).
Sothecoordinatesofothertwoverticesare
(8,−3)and(−2,1).
ans−
