Question

Prove that the diagonals of the parallelogram formed by 4 straight line are at right angles to each other.
\[L1=\sqrt{3}x+y=0\]
\[L2=x+\sqrt{3}y=0\]
\[L3=\sqrt{3}x+y=1\]
\[L4=x+\sqrt{3}y=1\]

Answer 1

i know one way by finding all the corner coordinates.

Answer 2

but we have to solve it using family of lines.

Answer 3

@mathslover @ganeshie8 @hartnn @Hero

Answer 4

can we say that : \(\large{\sqrt{3}{x}+y=0=\sqrt{3}{x}+y=1}\) ???

Answer 5

no

Answer 6

is \(\large{\sqrt{3}x+y=\sqrt{3}{x}+y}\)

Answer 7

yes

Answer 8

can u find angle between lines ?
do u need to use pair of lines equation ?
ax^2+2hxy+by^2 ??

Answer 9

given : \

L1 = \(\sqrt{3}{x+y=0}\)
L3 = \(\sqrt{3}{x}+y=1\)

How is it possible?

Answer 10

parallels

Answer 11

those are equatiion of lines

Answer 12

0 = 1?

Answer 13

yes

Answer 14

OK gotcha

Answer 15

Sorry :)

Answer 16

(0,0) is one corner as both adjacent lines pass thru origin (y = mx)
and (0,1) is another corner as other set of adjacent lines has same y-intercept = 1

Answer 17

so the slope of one diagonal is undefined

Answer 18

we need to prove slope of other diagonal = 0

Answer 19

is there any other method which does not involves finding the corners??

Answer 20

let me think.. .

Answer 21

a rectangle is a parallelogram but doesn't have perpendicular diagonals. could you revise your question.

Answer 22

a square and rhombus are also parallelogram

Answer 23

you can say for parallelogram them, parallelogram means rectangle and sheared rectangle too .... and square is a rhombus. perhaps you should have said for rhombus.

Answer 24

also your approach works ... if you are taking those 4 lines as 4 sides of rhombus then it should work fine.
also you could try taking 2 diagonals directly as diagonals will be bisecting two sides from on the point they meet.

Answer 25

i just want to know , how can we find the equations of diagonals from eq. of 4 sides , without finding the corner points??