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Q.

Two opposite vertices of a square are $− 1, 2$ and $3, 2$ . Find the coordinates of the other two vertices.

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a

(1, 0) and (1, 4)

b

(2, 0) and (4, 1)

c

(0, 1) and (1, 4)

d

(2, 0) and (2, 1)

answer is A.

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Detailed Solution

It is given that a square has opposite sides A(-1, 2), and C(3, 2).
Question Image

Question Image

Let B(x, y) be another vertex.
We shall use the property of square that all its sides are equal and it's two adjacent sides forms a right-angled triangle with the diagonal as it's hypotenuse.
We know all the sides of square are equal.
Therefore, we can write that AB=CB.
Using distance formula,
AB=

\sqrt{{(x + 1)}^{2} + {(y - 2)}^{2}}

CB=

\sqrt{{(x - 3)}^{2} + {(y - 2)}^{2}}

Now,
AB=CB
⇒

\sqrt{{(x + 1)}^{2} + {(y - 2)}^{2}} = \sqrt{{(x - 3)}^{2} + {(y - 2)}^{2}}

Cancelling the square roots,

{(x + 1)}^{2} + {(y - 2)}^{2} = {(x - 3)}^{2} + {(y - 2)}^{2}

{(x + 1)}^{2} = {(x - 3)}^{2 +} {(y - 2)}^{2} - {(y - 2)}^{2}

{(x + 1)}^{2} = {(x - 3)}^{2}

{(x + 1)}^{2} - {(x - 3)}^{2} = 0

x^{2} + 1 + 2 x - (x^{2} + 9 - 6 x) = 0

x^{2} + 1 + 2 x - x^{2} - 9 + 6 x = 0

8 x - 8 = 0

8 x = 8

x = \frac{8}{8}

x = 1

Let us join AC
Since all angles of square are

90^{°}

, ΔABC is a right-angled triangle.
Applying the Pythagoras theorem,

{AC}^{2} = {AB}^{2} + {BC}^{2}

{[{\sqrt{{(3 + 1)}^{2} + {(2 - 2)}^{2}}}^{}]}^{2} = {[\sqrt{{(x + 1)}^{2} + {(y - 2)}^{2}}]}^{2} + {[\sqrt{{(x - 3)}^{2} + {(y - 2)}^{2}}]}^{2}

{(4)}^{2} + {(0)}^{2} = {(x + 1)}^{2} + {(y - 2)}^{2} + {(x - 3)}^{2} + {(y - 2)}^{2}

16 = x^{2} + 1 + 2 x + y^{2} + 4 - 4 y + x^{2} + 9 - 6 x + y^{2} + 4 - 4 y

16 = 2 x^{2} - 4 x + 2 y^{2} - 8 y + 18

Putting the value of x as 1,

16 = 2 {(1)}^{2} - 4 (1) + 2 y^{2} - 8 y + 18

16 = 2 - 4 + 2 y^{2} - 8 y + 18

16 = 16 + 2 y^{2} - 8 y

16 - 16 = 2 y^{2} - 8 y

0 = 2 y^{2} - 8 y

0 = 2 y (y - 4)

y (y - 4) = 0

So, y = 0, 4 .
Hence, the correct option is 1.

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