CBSE Class 10 Answered
Answer : Given :If two opposite vertices of a square are ( 5,4 ) and ( 1,6)
To find : the coordinates of its remaining vertices
Let B (x, y).
We have AB = BC
Squaring both sides
=> AB2 = BC2
=> (x-5)2 +(y-4)2 = (x-1)2 +(y-6)2
=> x2 +25 -10x +y2 +16 - 8y = x2 +1 -2x + y2 +36 -12y
=> 4 - 8x +4y =0
=>2x - y =1 .........(1)
also AC2 = (5-1)2 + (4-6)2 = 16 + 4 = 20
In right triangle ABC, by Pythagoras theorem,
AC2 = AB2+BC2
=> AC2 = 2AB2
=> 20 = 2 ( (x-5)2 +(y-4)2 )
=> 20 = 2 ( x2 + 25 -10x+ y2 +16 -8y )
=> 10= x2 + 25 -10x+ y2 +16 -8y
=> x2 -10x+ y2 -8y + 31 =0
=> x2 -10x+ (2x+1)2 - 8(2x+1) + 31 =0 {using eq 1}
=> x2 -10x+ 4x2 + 1 + 4x -16x-8 + 31 =0
=>5 x2 -22x +24 =0
D = b2 -4ac
=> D = (-22)2 - 4(5)(24)
=> D = 484 - 480
=> D =4
=> D1/2 = 2
x = (-b+D1/2) / 2a , (-b - D1/2) / 2a
=> x = (22 +2) / 10 , (22-2) / 10
=> x = 2.4 , 2
=> y = 2x+1 , => y = 5.8 , 5
therefore the points are ( 2.4 , 5.8) and ( 2,5) Answer