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प्रश्न
Two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of other two
vertices.
उत्तर
The distance d between two points `(x_1,y_1)` and `(x_2,y_2)`
`d = sqrt((x_1- x_2)^2 + (y_1 - y_2)^2)`
In a square, all the sides are of equal length. The diagonals are also equal to each other. Also in a square, the diagonal is equal to `sqrt2` times the side of the square.
Here let the two points which are said to be the opposite vertices of a diagonal of a square be A(−1,2) and C(3,2).
Let us find the distance between them which is the length of the diagonal of the square.
`AC = sqrt((-1-3)^2 + (2 - 2)^2 )`
`= sqrt((-4)^2 +(0)^2)`
`= sqrt(16)`
AC = 4
Now we know that in a square,
The side of the square = `"Diagonal of the square"/sqrt2`
The side of the square = `2sqrt2`
Now, a vertex of a square has to be at equal distances from each of its adjacent vertices.
Let P(x, y) represent another vertex of the same square adjacent to both ‘A’ and ‘C’
`AP = sqrt((-1-x)^2 + (2 -y)^2)`
`CP = sqrt((3 - x)^2 + (2 - x)^2)`
But these two are nothing but the sides of the square and need to be equal to each other.
AP = CP
`sqrt((-1-x)^2 + (2 - y)^2) = sqrt((3 - x)^2 + (2 - y)^2)`
Squaring on both sides we have,
`AP = sqrt((-1-x)^2 + (2 - y)^2)`
`2sqrt(2) = sqrt((-1-1)^2 + (2 - y)^2)`
`2sqrt2 = sqrt((-2)^2 + (2 - y)^2)`
Squaring on both sides,
`8 = (-2)^2 + (2 - y)^2`
`8 = 4 + 4 = y^2 - 4y`
`0 = y^2 - 4y`
We have a quadratic equation. Solving for the roots of the equation we have,
`y^2 - 4y = 0`
y(y - 4) = 0
The roots of this equation are 0 and 4.
Therefore the other two vertices of the square are (1, 0) and (1,4)
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