Question

The lines joining the origin to the points of intersection of 2x^2+3xy-4x+1=0 and 3x+y=1 is given by __________

Answer 1

I tried solving the 2 equations first but it doesnt seem to be do the magic.

Answer 2

@ganeshie8 Any idea?

Answer 3

So, you want the line that goes through the origin (0,0) and the intersection point of the two given lines (x,y)?

Answer 4

the "lines"

Answer 5

Oh-ho, silly Austin... one is a quadratic!
Gimme a minute to ponder this, I will get right back to you!

Answer 6

Take your time :)

Answer 7

Do you have a graphing calculator?

Answer 8

yes http://www.wolframalpha.com/input/?i=2x%5E2%2B3xy-4x%2B1%3D0+and+3x%2By%3D1

Answer 9

ignore first 3 lines

Answer 10

But I am not allowed to use it in actual exam. So that doesnt help.

Answer 11

tell if i have done smthng wrong?

Answer 12

You can't use a TI-83/84 on an exam?

Answer 13

No calculators allowed. Besides I am looking for a solid answer.

Answer 14

@divu.mkr Can you explain? I am confused.

Answer 15

just took the equation of y from 1st equation, putted into 2nd ,got the locus of intersection points....then got the slope for d equation by diff..

Answer 16

What did you do in the 4th step?

Answer 17

can you specify by line no. from the top?

Answer 18

4th line.

Answer 19

ignoring first 3 lines then i just equated the values of y and not ignoring then i took the slope :D

Answer 20

Still confused. I got the quadratic 7x^2-x+1=0

Answer 21

if my algebra is a bit wrong then go by concept :P

Answer 22

Yeah But I dont know what to do after that.

Answer 23

slope!!!

Answer 24

How?

Answer 25

diff it

Answer 26

It gives the slope of what?

Answer 27

Ok, we have two different equations.

One is of a hyperbola:
\(2x^2+3xy-4x+1=0 \)

The other is of a line:
\(3x+y=1\)

We can find the points of intersection through a series of steps.

1) Solve the equation of the line for x.

\(3x+y=1\)
\(3x=-y+1\)
\(\displaystyle x=-\frac{y}{3}+\frac{1}{3}\)

2) Substitute that into the equation for the hyperbola.

3) Expand and group like terms and rewrite the equation.

4) Solve the quadratic equation for y to obtain two solutions.

5) We now substitute the values of y already obtained into the equation we have for x.

Answer 28

^very long process. I am sure there is a better way out.

Answer 29

thats where i made a mistake....:D
PS- get x by solving 7x^2+x-1=0

Answer 30

hmm....It is a long process. I am thinking of any easier method.

Answer 31

i think there is.. but i think i have to remind my notes...

Answer 32

ok

Answer 33

2x^2+3xy-4x+1=0 ------(1)
3x+y=1 -------(2)

they can have two intersecting points,
so the equation for pair of lines through origin wud be :-
2x^2+3xy-4x(3x+y)+1(3x+y)^2=0 ---------------(3)

Answer 34

proof is trivial :-
say A(m, n) is one of the intersection point,
then A satisfies (1) and (2) << GIVEN.

clearly it satisfies (3) also.

Answer 35

also, (3) satisfies (0, 0)
so origin, and the two intersection points of (1) and (2) lie on the two straight lines represented by (3)

Answer 36

Ah! That makes all sense. So you just homogenized the given conditions isnt it?

Answer 37

exactly !
i never learned these in my highschool/college,
recently i happened to learn to these while working on few problems of @samigupta8
she did many varieties of these problems and knows much better than me... the internals of theorems involved and stuff behind these...

Answer 38

Ah thanks a lot ..Now I have a clear idea on how to do these.

Answer 39

np :)

Answer 40

A solution using Mathematica 8 Home Edition for the calculations is attached.

Answer 41

Thanks :)