Question: The opposite angular points of a square are $ (3,\ 4) $ and $ (1,\ -\ 1) $ . Then the co-ordinates of other two points are [Roorkee 1985]
Options:
A) $ D( \frac{1}{2},\frac{9}{2} ),B( -\frac{1}{2},\frac{5}{2} ) $
B) $ D( \frac{1}{2},\frac{9}{2} ),B( \frac{1}{2},\frac{5}{2} ) $
C) $ D( \frac{9}{2},\frac{1}{2} ),B( -\frac{1}{2},\frac{5}{2} ) $
D)None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Obviously, slope of $ AC=5/2 $ . Let m be the slope of a line inclined at an angle of $ 45^{o} $ to AC, then $ \tan 45^{o}=\pm \frac{m-\frac{5}{2}}{1+m.\frac{5}{2}}\Rightarrow m=-\frac{7}{3},\frac{3}{7} $ .Thus, let the slope of AB or DC be 3/7and that of AD or BC be $ -\frac{7}{3} $ . Then equation of AB is $ 3x-7y+19=0 $ . Also the equation of BC is $ 7x+3y-4=0 $ . On solving these equations, we get, $ B( -\frac{1}{2},\frac{5}{2} ) $ . Now let the coordinates of the vertex D be (h, k). Since the middle points of AC and BD are same, therefore $ \frac{1}{2}( h-\frac{1}{2} )=\frac{1}{2}(3+1)\Rightarrow h=\frac{9}{2} $ , $ \frac{1}{2}( k+\frac{5}{2} )=\frac{1}{2}(4-1) $
Þ $ k=\frac{1}{2} $ . Hence, $ D=( \frac{9}{2},\frac{1}{2} ) $ .
BETA
