Math, asked by shahraghunath405, 14 hours ago

ABCD is a square of with the co-ordinates of two opposite vertices A and C are (2,2) and (6,6) respectively. find the co-ordinates of B and D.

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Answers

Answered by 4971jeevanthika
2

Answer:

Solution

Let ABCD is a square where two opposite vertices are A(−1,2)andC(3,2).

Let B(x,y) and D(x1.y1) 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

⇒x2+2x+1+y2+4−4y=9+x2−6x+4+y2−4y

⇒2x+5=13−6x

⇒2x+6x=13−5

⇒8x=8

⇒x=1

In △ABC,∠B=90

Step-by-step explanation:

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Answered by PURVA0246
1

Answer:

Let ABCD is a square where two opposite vertices are A(−1,2)andC(3,2).

Let B(x,y) and D(x1.y1) 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

⇒x2+2x+1+y2+4−4y=9+x2−6x+4+y2−4y

⇒2x+5=13−6x

⇒2x+6x=13−5

⇒8x=8

⇒x=1

In △ABC,∠B=90∘  [all angles of square are 90

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