Maths Challenge! (1 Viewer)

[Real Madrid]

Find the equation of the locus of a point that moves so that it is equidistant from the line

4x-3x+2=0
3x+4y-7=0

Show working

 

[Timothy.Siu]

Find the equation of the locus of a point that moves so that it is equidistant from the line

4x-3x+2=0
3x+4y-7=0

let P be a point (x,y)
perpendicular distance from both the lines is

D= (4x-3y+2)/root(x^2+y^2)

D= (3x+4y-7)/root(x^2+y^2)

4x-3y+2=3x+4y-7
x-7y+9=0 dunno if thats correct

 

[lolokay]

^ you forgot the absolute values
it would be:
4x-3y+2=3x+4y-7
x-7y+9=0

or
4x - 3y + 2 = -(3x + 4y - 7)
7x + y - 5 = 0

 

[tacogym27101990]

Find the equation of the locus of a point that moves so that it is equidistant from the line

4x-3x+2=0
3x+4y-7=0

let P be a point (x,y)
perpendicular distance from both the lines is

D= (4x-3y+2)/root(x^2+y^2)

D= (3x+4y-7)/root(x^2+y^2)

4x-3y+2=3x+4y-7
x-7y+9=0 dunno if thats correct

right idea, just got the formula wrong
but because of the nature of the question you still got the correct answer

shoulde be |4x-3y+2|/5=|3x+4y-7|/5 ( 5=sqr(3^2+4^2))

so 4x-3y+2=3x+4y-7 or 4x-3y+2 = -3x-4y+7
x-7y+9=0 or 7x+y-5=0

and im not sure why you call this a maths challenge??
sure oyu just couldnt do it?

 

[shaon0]

Find the equation of the locus of a point that moves so that it is equidistant from the line

4x-3x+2=0
3x+4y-7=0

Show working

can you stop naming all your threads with challenge or some other superlative when the question is not very hard or a challenge.
Thanks

 

[lyounamu]

can you stop naming all your threads with challenge or some other superlative when the question is not very hard or a challenge.
Thanks

QFT

 

[5647382910]

right idea, just got the formula wrong
but because of the nature of the question you still got the correct answer

shoulde be |4x-3y+2|/5=|3x+4y-7|/5 ( 5=sqr(3^2+4^2))

so 4x-3y+2=3x+4y-7 or 4x-3y+2 = -3x-4y+1
x-7y+9=0 or 7x+y-5=0

and im not sure why you call this a maths challenge??
sure oyu just couldnt do it?

wiht ur second answer how come |3x + 4y -7| = -3x -4y + 1 (im talking about the -7 changing to 1)? and also with the absolute values why did you do both negative and positive for the RHS and not the left (|4x - 3y + 2|)?

 

[lolokay]

wiht ur second answer how come |3x + 4y -7| = -3x -4y + 1 (im talking about the -7 changing to 1)? and also with the absolute values why did you do both negative and positive for the RHS and not the left (|4x - 3y + 2|)?

probs a typo

and because taking the negative of both sides is the same of having the positive of both sides, and taking the negative of only one side is the same as taking the negative of only the other (multiply each side by -1)

 

[Timothy.Siu]

? and also with the absolute values why did you do both negative and positive for the RHS and not the left (|4x - 3y + 2|)?

because u dont need to, its the same
e.g. |x+2|=|2x-1|
u only need to do 2 cases not 4
as -(x+2)=2x-1 is the same as x+2=-(2x-1)

 

[5647382910]

because u dont need to, its the same
e.g. |x+2|=|2x-1|
u only need to do 2 cases not 4
as -(x+2)=2x-1 is the same as x+2=-(2x-1)

ok i see.. thanks

 

[tacogym27101990]

yeah sorry was just a typo