Question

(0, –1) and (0, 3) are two opposite vertices of a square. The other two vertices are.

• A. (0, 1), (0, –3)
• B. (3, –1) (0, 0)
• C. (2, 1), (–2, 1)
• D. (2, 2), (1, 1)

Answer 1

Answer: (C)

(2, 1), (–2, 1)

Answer 2

Length of diagonal = 4

Now , $A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 }$

$A C ^ { 2 } = 2 A B ^ { 2 } \Rightarrow 8 = A B ^ { 2 }$

$A B = B C = 2 \sqrt { 2 }$

Now, let $B ( x , y )$; $\therefore A B ^ { 2 } = B C ^ { 2 }$

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

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

⇒ $y = 1 ; \therefore x ^ { 2 } + ( 2 ) ^ { 2 } = 8 ; \Rightarrow x ^ { 2 } = 4 \Rightarrow x = \pm 2$

$\therefore$ other vertices are (2, 1),(–2, 1).