Plz help me with a coordinate geometry sum..Given O(0,0),P(1,2),S(-3,0). P divides OQ in the ratio 2:3 and OPRS is a parallelogram. Find: i) Coordinate of Q ii) Coordinate of R iii) The ratio in...

Plz help me with a coordinate geometry sum..

Given O(0,0),P(1,2),S(-3,0).
P divides OQ in the ratio 2:3 and OPRS is a parallelogram. Find:
i) Coordinate of Q
ii) Coordinate of R
iii) The ratio in which RQ is divided by y axis.

Asked on by itssnigdha

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If P divides OQ in the ratio 2:3, that means OP=2 and PQ=3 and OQ=2+3=5

This thing is helping us to find out the coordinates for Q.

The formula for the length of PQ is:

PQ=sqrt[(xQ-xP)^2+(yQ-yP)^2]

PQ^2=[(xQ-1)^2+(yQ-2)^2]

9=xQ^2-2xQ+1+yQ^2-4yQ+4

xQ^2-2xQ+yQ^2-4yQ=9-5

xQ^2-2xQ+yQ^2-4yQ=4

OP^2=xQ^2+y^2

25=xQ^2+y^2

xQ^2-2xQ+yQ^2-4yQ=4, but 25=xQ^2+yQ^2

-2xQ-4yQ+25=4

-2xQ-4yQ=-21

xQ=(21-4yQ)/2

We'll substitute xQ into 25=xQ^2+yQ^2

25=[(21-4yQ)/2]^2+yQ^2

100=441-168yQ+20yQ^2

20yQ^2-168yQ+341=0

yQ=[168+sqrt(168^2-4*20*341)]/2*20

yQ=(168+32)/40

yQ=(200)/40=5

xQ=(21-4*5)/2

xQ=1/2

For finding the coordinates for R, you have to respect the given facts that QPSR is a parallelogram, meaning that the opposite sides are parallel and equal, so RS=PQ=3.

For this reason, you have to calculate the following relations:

RS=sqrt[(xR-xS)^2+yR^2]

9=[(xR+3)^2+yR^2]

RQ=SP=sqrt{[(1/2)-xR]^2+(5-yR)^2}

where SP=sqrt[(1+3)^2+(2-0)^2]

SP=sqrt(16+4)

SP=sqrt(20)

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neela | High School Teacher | (Level 3) Valedictorian

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P divides OQ in 2:3 ratio. Let (x1,y1) be the the coordinates of

Therefore, OP/OQ = 2/(2+3). So, {(Px -0)^2+(Py-0)^2] /[(Qx-0)^2+Qy-0)^2] =2/5 Or

=(1^2+2^2)/{x1^2+y1^2) = (2^2/5^2) =4/25 or

5*25 = 4(x1^2+y1^2)  or but x1 and y1 are on the line (0,0) and (1,2) which is y=2x. Threfore, 125 =4(x1^2+(2x1)^2) = 20x1^2 . So x1 ^2 = 125/20 =25/4 or x1 = +5/2 or x1 =-5/2 and y coordinate = 2x1 = 5 or -5. Therefore, Q is at (5/2,5) or at(-5/2, -5).

Cordinate of R :OS//OR. Therefore, yR =2.But y= 2x is the equation of OR. Threfore, Y = 2x+k is the equation of SR and passes through S(-3,0). So,  0 = 2(-3)+k or k = -6.Therefore, The equation of SR is y =2x+6. But Ry = 2. So Rx  is got by putting y=2 in y=2x+6 . Therefore, 2=2x +6 or 2x= 2-6. So x=-4/2 =-2. Therefore R(x,y) = (-2,2).

Equation of RQ  is  the line joining (-2,2) to (5/2, 5) which is

y-2 = (5-2)/(5/2--2)[x--2) or

y-2 = [3/(4.5))](x+2) = (2/3)(x+2) or

3y-6=2x+4  or 3y = 2x+10 which is intersected by y axis at  putting x=0 and y =( 2*0+10)/3 or at (0,4).

Therefore. RP/RQ =sqrt{ [(0--2)^2+(4-2)^2]/(5/2-0)^2+(5-4)^2}

= sqrt{(8/(7.25)}

 

 

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