Question

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

Answer 1

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)

Answer 2

Answer:

hence, the co-ordinates of P, $(1,\frac{-3}{2})$ and $(3,\frac{-1}{2})$

Step-by-step explanation:  

QS = 2PR

distance between any two points $(x_1,y_1) \text{and} (x_2,y_2)$ is calculated by $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

so, distance between (1,1) and (2,-1) is ;

$\sqrt{(2-1)^{2}+(-1-1)^{2}}$

$\sqrt{(1)^{2}+(-2)^{2}}$

$\sqrt{(1)+4}$

$\sqrt{5}$

Since, QS = 2PR

SM=SQ =$\sqrt{5}$ and PM = PR = $\frac{\sqrt{5}}{2}$

Slope of SM is calculated by using slope formula

$m_1=\frac{y_2-y_1}{x_2-x_1}$

$m_1=\frac{-1-1}{2-1}$

$m_1=\frac{-2}{1}$

$m_1=-2$

since, diagonals are perpendicular to each other then it must have product of their slope is -1.

let $m_2$ be the slope of line PR

$m_1\times m_2 = -1$

$m_2 =\frac{-1}{m_1}$

$m_2 =\frac{-1}{-2}$

$m_2 =\frac{1}{2}$

Equation of line PR

$y + 1 =\frac{1}{2}(x-2)$

multiply both the sides by 2, in above equation

$2(y + 1) =x-2$

$2y + 2 =x-2$

subtract 2 from both the sides,

$2y=x-4$

divide both the sides vy 2, in above equation

$y=\frac{x-4}{2}$

let x be 'h'

then

$(h,\frac{h-4}{2})$

distance between PM ;

$\sqrt{(h-2)^{2}+(\frac{h-4}{2}+1)^{2}}}$=$\frac{\sqrt{5}}{2}$

squaring both the sides on above equation

$(h-2)^{2}+(\frac{h-2}{2})^{2}}$=$\frac{5}{4}$

$\frac{5(h-2)^{2}}{4}}$=$\frac{5}{4}$

$(h-2)^{2}=1$

after simplifying the above , we get

h = 1 or 3

if h = 1 then $(1,\frac{1-4}{2})$ =$(1,\frac{-3}{2})$

if h = 3 then $(3,\frac{3-4}{2})$ =$(3,\frac{-1}{2})$

hence, the co-ordinates of P, $(1,\frac{-3}{2})$ and $(3,\frac{-1}{2})$