if pqrs is a rhombus then show that pr biscuts<pand<r
Answers
Answer:
(2 , 3/2 ) & (0 , 1/2)
Step-by-step explanation:
If PQRS is a Rhombus.Its diagonals PR and QS intersect at the point M and satisfy QS=2PR.If the co-ordinates of S and M are (1,1) and (2,-1) respectively.Find the co-ordinates of P
Diagonal of rhombus bisect each other at 90°
QS = 2PR
=> SM = 2 PM & SM ⊥ PM
co-ordinates of S and M are (1,1) and (2,-1)
Slope of SM = (1 - (-1))/(1-2) = 2/-1 = -2
Multiplication of Slope of Perpendicular line is - 1
slope of PM = 1/2
y = x/2 + c
putting coordinates of M
1 = 1/2 + c
=> c = 1/2
y = x/2 + 1/2
2y = x + 1
=> x = 2y - 1
SM² = (2-1)² + (-1-1)² = 5
PM = SM/2 => PM² = SM²/4 = 5/4
P(x ,y)
PM² = (x -1)² + (y-1)² = 5/4
putting x = 2y-1
=> (2y -1 -1)² + (y -1)² = 5/4
=> 4(y-1)² + (y -1)² = 5/4
=> 5(y-1)² = 5/4
=> (y-1)²= 1/4
=> y -1 = ± 1/2
=> y = 3/2 , 1/2
x = 2y -1 = 2 , 0
(2 , 3/2 ) & (0 , 1/2)