Question

find the locus of a point P that moves such that the distance of P from the lines 3x - 4y + 1 = 0 and 12x +5y + 3 = 0 is in the ratio 3:1

Answer 1

there is a formula for distance from a point to a line , you might be able to use that

Answer 2

oh, you mean this one

\[d = \frac{\left| Ax + By + C \right|}{\sqrt{A^2 + B^2}}\]

Answer 3

the formula is
distance ( xo, yo ) ( ax + by + c = 0 ) = | axo + byo + c | / sqrt(a^2 + b^2 )

Answer 4

yes :)

Answer 5

where the point is P(x, y)

Answer 6

so this might work
solve
| 3x - 4y + 1 | / sqrt( 3^2 + 4^2 ) = 3 * | 12x + 5y + 3 | / sqrt( 12^2 + 5^2 )

Answer 7

the locus is a pair of intersecting lines

Answer 8

so its then

\[\frac{13(3x - 4y + 1)}{3} = \frac{5(12x + 5y + 3)}{1}\]

or

\[(39x - 52y + 13) = 3(50x + 25y + 15)\]

so then you would say if the point is between the lines then use 3 and collect line terms

then is the lines are in the same side of the point you would use -3 since its the external division in a ratio

is that what you're saying..?

Answer 9

oops should read

\[(39x - 52y + 13) = 3(60x + 25y + 15)\]

Answer 10

almost, there are two cases because of the absolute value

Answer 11

oh i think you considered both cases

Answer 12

so what I wrote is what you are saying.... the point is between the lines...

and the lines can be on the same side of the point....

Answer 13

right

Answer 14

wow... thanks for the help..

Answer 15

also it would be nice to be able to test this , plug in points

Answer 16

maybe you you make a geogebra applet... so show it

Answer 17

yeah that would be cool :)

Answer 18

I'll ask my teacher if he can do it

Answer 19

I read about another side so posted it on there... and people gave an elipseand circle as the answer... I knew that didn't look right

Answer 20

do you have a link to that

Answer 21

it was study pool... tutors can get paid...

Answer 22

its difficult to test because I get fractions when I solve for y

Answer 23

why solve for y... its a locus for the values in the point keep changing and are described by the 2 linear equations...

Answer 24

i want to plot all three lines and test it

Answer 25

all that would acheive is showing a specific P

Answer 26

right

Answer 27

and you can move P around

Answer 28

I'd need to look at it and the point P in the solution line can be moved

Answer 29

anyway.... that's my Year 11 homework done... thanks

Answer 30

:)

Answer 31

I just tested it for the point P ( 1, -277/23) , and it works. I used the negative line
http://www.wolframalpha.com/input/?i=simplify+-%28+3x+-+4y+%2B+1+%29+%2F5+%3D+3+%28+12x+%2B+5y+%2B+3+%29+%2F13

Answer 32

distance [( 1, -277/23) , ( 3x -4y + 1 = 0 ) ] = 240/23
distance [(1,-277/23) , ( 12x + 5y + 3 = 0 ) ] = 80/23