Home » Ask & Discuss » Mathematics. » Analytical Geometry « Back to Discussion

Analytical Geometry

Blazing goIITian

Joined:	14 Aug 2007
Post:	910

29 Oct 2007 22:18:48 IST

0 People liked this

545

kindly help

None

1.two sides of a rhombus lying in the 1st quadrant are given by 3x-4y=0 and 12x-5y=0.if the length of the longer diagonal is 12,find the equations of the other two sides of the rhombus.

ans.3x-4y= -180/square root130 & 12x-5y=468/square root130

2.a straight line passes through a fixed point (h,k).find the locus of foot of perpendicular on it drawn from origin.

ans.x²+y²-hx-ky=0

please explain the steps properly.rates guaranteed.

Share this article on:

Comments (6)

Ankur Limaye

Cool goIITian

Joined: 7 Jul 2007

Posts: 98

29 Oct 2007 22:43:16 IST

Like 0 people liked this

Dude .....

Check the first question .....

Both the lines 3x - 4y = 0 and 12x - 5y = 0 intersect at origin .....

(How come it may be a rhombus .....

)

Nadeem

Blazing goIITian

Joined: 25 Aug 2007

Posts: 521

29 Oct 2007 23:17:50 IST

Like 1 people liked this

1)
2 Adjacent sides of the rhombus are 3x - 4y = 0 and 12x - 5y = 0
Let the other sides be 3x - 4y = a and 12x - 5y = b

First find the equation of the angle bisectors of the given lines.
|3x-4y|/5=|12x-5y|/13
9x = 7y or 9x+7=0

Now let 3x - 4y = a and 12x - 5y = b meet at ( p,q)

(p,q) = ( p, 9p/7 ) or (p,-9p/7) as it lies on the angle bisector.

Length of longer diagonal =12

Therefore ( p-0)^2 + ( 9p/7 -0)^2 = 12^2
p= 84/rt130 (*)
q=108/rt130

Now substitute this in 3x - 4y = a and 12x - 5y = b and you can get the given answers.

(*) If -ive sign is taken , it will not be the longer diagonal

rini s

Blazing goIITian

Joined: 31 May 2007

Posts: 495

29 Oct 2007 23:18:25 IST

Like 1 people liked this

(1)

the ques is correct...

the two lines pass from origin, so one end of longer digonal is origin.

now, the equation of digonal is the angle bisector of the two given lines..

| 3x - 4y|/5 = |12x-5y|/13..( take that equation in which slope is positive coz digonal is in 1st quad.)

the equation cums out to be.. x=(7/9)y

and length of digonal i.e. ( x²+y²)^1/2 = 12

solving both equations.. we get coordinates of end of digonal = (7,9)

now the slope of other two sides is equal to the two given sides respectively..

so equation of sides.. y=(3/4)x + c₁ and y=(12/5)x + c₂

and both lines pass through (7,9) ...{end point of digonal }

so putting x=7 and y=9 in both the equations we get value of c1 and c2 anh hence the required equation cums out ...

rini s

Blazing goIITian

Joined: 31 May 2007

Posts: 495

29 Oct 2007 23:27:20 IST

Like 2 people liked this

(2)

let the straight line is... (y-k) = m(x-h)

so m= (y-k)/(x-h)

noe the line perpendicular to this line have slope.. = - ( x-h)/(y-k)

so equation of the perpendicular:

y = - [(x-h)/(y-k) ] x

=> x²+ y² - hx - ky =0.................(answer

)

Nadeem

Blazing goIITian

Joined: 25 Aug 2007

Posts: 521

29 Oct 2007 23:50:12 IST

Like 1 people liked this

coordinate of end of diagonal is not (7,9)

Arshdeep Singh Malhan

Blazing goIITian

Joined: 14 Aug 2007

Posts: 910

30 Oct 2007 13:14:40 IST

Like 0 people liked this

thnx everybody for your assistance

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team