Question

Any two coordinates of a square are (4, 4) and (4, –5). The other two coordinates cannot be

• A. $(\frac{-1}{2},\frac{-1}{2})and(\frac{17}{2},\frac{1}{2})$
• B. $(\frac{-1}{2},\frac{-1}{2})and(\frac{17}{2},\frac{-1}{2})$
• C. (–5, 4) and (–5, –5)
• D. (13, 4) and (13, –5)

Answer 1

The correct option is A $(\frac{-1}{2},\frac{-1}{2})and(\frac{17}{2},\frac{1}{2})$

Let the given two coordinates of the square be A(4, 4) and B(4, –5).


$AB=\sqrt{(4-4{)}^2+(4+5{)}^2}$

= 9 units

Case 1: When the vertices A and B are adjacent

Since, A, and B are 9 units apart, therefore, 2 squares can be made out of it of sides 9 units.

So, 9 units from A(4, 4) will be H(–5, 4) to the left and 9 units from the point B(4, –5) will be G(–5, –5) to the left.

Hence, square ABGH is formed with coordinates A(4, 4), B(4, –5), G(–5, –5) and H(–5, 4).

Similarly, 9 units from A and B to the right will make points E(13, 4) and F(13, –5) and a square ABFE is formed.


Case 2: When the vertices A and B are opposite, i.e., AB is the diagonal

Let the side of the square be a, then

$a^2+a^2=9^2$ (By Pythagoras Theorem)

$\Rightarrow 2a^2=9^2$

$\Rightarrow a^2=\frac{9^2}{2}$

$\Rightarrow a=\frac{9}{\sqrt{2}}$

Now, let D(x, y) be a vertex of 3rd square.

Then, $AD=\sqrt{(4-x{)}^2+(4-y{)}^2}$

$\Rightarrow \frac{9}{\sqrt{2}}=\sqrt{(4-x{)}^2+(4-y{)}^2}$ ...(i)

Similarly,

$DB=\sqrt{(4-x{)}^2+(-5-y{)}^2}$

$\Rightarrow \frac{9}{\sqrt{2}}=\sqrt{(4-x{)}^2+(-5-y{)}^2}$ ...(ii)

Also, AD = BD

$\Rightarrow A{D}^2=B{D}^2$

$(4-x{)}^2+(-5-y{)}^2=(4-x{)}^2+(4-y{)}^2$

$\Rightarrow 25+y^2+10y=16+y^2-8y$

$\Rightarrow 18y=-9$

$\Rightarrow y=\frac{-1}{2}$ ...(iii)

Putting (iii) in (i),

$\frac{9}{\sqrt{2}}=\sqrt{(4-x{)}^2+{(4+\frac{1}{2})}^2}$

$\Rightarrow \frac{81}{2}=(4-x{)}^2+{\left(\frac{9}{2}\right)}^2$

$\Rightarrow (4-x{)}^2=\frac{81}{2}-\frac{81}{4}$

$\Rightarrow (4-x{)}^2=\frac{162-81}{4}$

$\Rightarrow (4-x{)}^2=\frac{81}{4}$

$\Rightarrow 4-x=\frac{9}{2}$

$\Rightarrow x=4-\frac{9}{2}$

$\Rightarrow x=\frac{-1}{2}$

Thus, the coordinates of D are $(\frac{-1}{2},\frac{-1}{2}).$

Similarly, the coordinates of vertex C will be $(\frac{17}{2},\frac{-1}{2}).$



Thus, ACBD will be the 3rd square.

Hence, $(\frac{17}{2},\frac{1}{2})$ will not make a square with the given coordinates.

Hence, the correct answer is option (a).