opposite vertices of a rectangle are (-1,-2) and (2,-4).Find the coordinates of other vertices
Answers
Let's draw a figure of a square with the two opposite vertices (-1, 2) and (3, 2),
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices
Let ABCD be a square having known vertices A (- 1, 2) and C (3, 2) respectively.
Let B(x₁, y₁) and D(x₂, y₂) be the two unknown vertex
We know that the sides of a square are equal to each other.
Therefore, AB = BC
By Using Distance formula on AB = AC with A (- 1, 2), B(x₁, y₁) and C (3, 2)
√ [(x₁ - (-1))2 + (y₁ - 2)2] = √ [(x₁ - 3)2 + (y₁ - 2)2]
x₁2 + 2x₁ + 1 + y₁2 - 4y₁ + 4 = x₁2 + 9 - 6x₁ + y₁2 + 4 - 4y₁ (By Simplifying & Transposing)
8x₁ = 8
x₁ = 1
We know that in a square, all interior angles are 90 degrees.
In ΔABC
AB2 + BC2 = AC2 [By Pythagoras theorem]
Step-by-step explanation:
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)