15. (-4,-1), (-1,-1), (-1, 2) and (I, m) are the vertices of a square. Find I, m and the length of the diagonal of the square.
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Answers
Answer:
(-4,-1), (-1,-1), (-1, 2) and (I, m) are the vertices of a square. Find I, m and the leyou are find the answer is 4 - 4 and -1
Vivah find the answers is - 4 - 1
ANSWER
Let ABCD is a square where two opposite vertices are A(−1,2)andC(3,2).
Let B(x,y) and D(x
1
.y
1
) ids the other two vertices.
In Square ABCD
AB=BC=CD=DA
Hence AB=BC
⇒
(x+1)
2
+(y−2)
2
=
(3−x)
2
+(2−y)
2
[by distance formula]
Squaring both sides
⇒(x+1)
2
+(y−2)
2
=(3−x)
2
+(2−y)
2
⇒x
2
+2x+1+y
2
+4−4y=9+x
2
−6x+4+y
2
−4y
⇒2x+5=13−6x
⇒2x+6x=13−5
⇒8x=8
⇒x=1
In △ABC,∠B=90
∘
[all angles of square are 90
∘
]
Then according to the Pythagorean theorem
AB
2
+BC
2
=AC
2
As AB=BC
∴2AB
2
=AC
2
⇒2(
x+1)
2
+(y−2)
2
)
2
=(
(3−(−1))
2
+(2−2)
2
)
2
⇒2((x+1)
2
+(y−2)
2
)=(3+1)
2
+(2−2)
2
⇒2(x
2
+2x+1+y
2
+4−4y)=(4)
2
Put the value of x=1
⇒2(1
1
+2×1+1+y
2
+4−4y)=16
⇒2(y
2
−4y+8)=16
⇒2y
2
−8y+16=16
⇒2y
2
−8y=0
⇒2y(y−4)=0
⇒(y−4)=0
Hence y=0 or 4.
As diagonals of a square are equal in length and bisect each other at 90
∘
Let P is the midpoint of AC
∴CO-ordinates of P=(
2
3−1
,
2
2+2
)=(1,2)
P is also the midpoint of BD
then co-ordinates of mid-point of BD=co-ordinates of P
⇒(x
1
,y
1
)=1,2
∴x
1
=1,y
1
=2
Then other two vertices of square ABCD are (1,0) or (1,4) and (1,2).
hope you understand
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