Please solve...The two opposite vertices of a square are (-1,2) and (3,2). Find the coordinates of the other two vertices........Please show steps.....

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Gopal Mohanty answered this

Hi!

Here is the answer to your question.

So, x₁ + x₂ = 2 and y₁ + y₂ = 4

When x₁ = 1, x₂ = 1

When y₁ = 0, y₂= 4

When y₁ = 4, y₁ = 0

Coordinates of D are (1, 4) or (1, 0)

Thus, the other two vertices are (1, 0) or (1, 4)

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Sakti Ghosh answered this

Mid point of the diagonal joining these vertices is ( 2 , 2 ) and length of the diagonal is 4 units. In square , diagonals bisect eachother perpendicularly. Also the diagonal joining the two given points is parallel to x-axis. Therefore the other diagonal will be parallel to y-axis and of length 4 units with its mid-point at ( 2 , 2 ). Therefore the other two vertices are ( 2 , 0 ) and ( 2 , 4 ).

Sakti Ghosh answered this

Extremely Sorry for the typographical mistake. Mid point will be ( 1,2)and the vertices will be ( (1,0) and ( 1,4)

Maryam Ruquia answered this

Lets say that B(-1,2) and D(3,2).. let C be (x,y)

BC=DC >> BC²=DC²

>> (x+1)²+ (y-2)² = (3-x)² + (2-y)² [BY USING DISTANCE FORMULA]

>> (x²+1+2x) + (y²+4 - 4y) = (x²+9-6x) + (y²+4-4y) [A²+B²-2AB = (A-B)²]

>> x²+1+2x + y²+4 - 4y = x²+9-6x + y²+4-4y [CANCELATION]

>> 5+2x-13-6x = 0

>> -8-4x = 0 >> x= -2

Maryam Ruquia answered this

Now, BD= 4 units [DISTANCE FORMULA]

O= (x1+x2/2, y1+y2/2) = (1,2)

>> OC²= 1/2 (4)² = 8 units

>> (-2-1)² + (y-2)² = 8

>> 9+ y²+ 4 - 4y - 8 =0 >> y²-4y+6= 0 >> y= 0 or 4

>> THEREFORE THE TWO VERTICES ARE (-2,0) and (-2,4).

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Sakti Ghosh answered this

When BD = AC = 4 units , how OC² = 8 ?

Maryam Ruquia answered this

BD=AC=4units; OC= 1/2 AC (its a square) >> OC²= 4units

>> (-2-1)² + (y-2)² = 4

>> 9+ y²+ 4 - 4y -4 =0

>> y²-4y +9 = 0

SOLVE THE EQUATION YOU'L GET THE ANSWER.....MAYBE!!! i'm not sure...if it's wrong then remember.... "I'm still LEARNING!!" lol :)