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  • The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the        other two vertices.

Q4   The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the
       other two vertices.

Answers (1)

best_answer

From the figure:

We know that the sides of a square are equal to each other.

Therefore, AB = BC

So,

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

Squaring both sides, we obtain

\implies (x-1)^2+(y-2)^2 = (x-3)^2+(y-2)^2

Now, doing \left ( a^2-b^2 = (a+b)(a-b) \right )

We get

\implies (x-1+x-3)(x-1-x+3) = 0

Hence x = 2.

Applying the Pythagoras theorem to find out the value of y.

AB^2+BC^2 = AC^2

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

\Rightarrow \left (\sqrt{1+(y-2)^2} \right )^2 + \left (\sqrt{1+(y-2)^2} \right )^2 = \left (\sqrt{16} \right )^2

\Rightarrow \left ({1+(y-2)^2 \right ) + \left (1+(y-2)^2 \right ) = 16

\Rightarrow (y-2)^2 = 7

Posted by

Divya Prakash Singh

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