Coordinate Geometry Question 129
Question: (0, -1) and (0, 3) are two opposite vertices of a square. The other two vertices are
[Karnataka CET 2005]
Options:
A) (0, 1), (0, -3)
B) (3, -1) (0, 0)
C) (2, 1), (-2, 1)
D) (2, 2), (1, 1)
Show Answer
Answer:
Correct Answer: C
Solution:
Length of diagonal = 4 Now, $ AC^{2}=AB^{2}+BC^{2} $
$ AC^{2}=2AB^{2}\Rightarrow 8=AB^{2} $
$ AB=BC=2\sqrt{2} $ Now, let $ B(x,y) $ ;
$ \therefore AB^{2}=BC^{2} $
therefore $ {{(x-0)}^{2}}+{{(y+1)}^{2}}={{(x-0)}^{2}}+{{(y-3)}^{2}} $
$ x^{2}+y^{2}+2y+1=x^{2}+y^{2}-6y+9 $
therefore $ y=1;\therefore x^{2}+{{(2)}^{2}}=8;\Rightarrow x^{2}=4\Rightarrow x=\pm 2 $
$ \therefore $ other vertices are (2, 1),(-2, 1).
BETA
