Parametrics Q (1 Viewer)

[shaon0]

1. If P(x,y) is any point on the line y=4x+3 show that the midpoint M of OP has the co-ordinates M(x/2,0.5(4x+3)). Hence find the locus of M
...i have done the 1st part i just don't know how to find the locus.

2. P(x,y) is a variable point on the line 2x-3y+6=0 and the pint Q divides OP in the ration 3:2. Show that the co-ordinates of Q are (3x/5, 2/5(x+3)). Hence find the locus of Q.
....I know what formula to use to find the co-ordinates of Q. just i don't know where to get the data so i can input it into the formula.

3. Show that if P(x1,y1,z1) and Q(x2,y2,z2) are two points in three dimensional space,
then (PQ)^2= (x2-x1)^2+(y2-y1)^2+(z2-z1)^2
Need help. Thanks

 

[lolokay]

1. let the x coordinate be x₁
then 2x₁ = x
y₁ = 0.5(8x₁ + 3)

2. 5/3x₁ = x
y₁ = 2/5(5/3 x₁+3)


3. just use pythagoras

 

[shaon0]

for 3. i can't use pythagoras can i because pythagoras is on a 2 dimensional plane, this is three dimensional.

 

[shaon0]

1. let the x coordinate be x₁
then 2x₁ = x
y₁ = 0.5(8x₁ + 3)

2. 5/3x₁ = x
y₁ = 2/5(5/3 x₁+3)


3. just use pythagoras

your part 2. does not make any sense.

 

[lolokay]

you've never used pythag in 3-d?
say you have 3 lines; x, y and z and you want to find the diagonal xyz
then for the xy diagonal (xy) xy² = x² + y²
and for the the xyz diagonal (xyz) xyz² = xy² + z²
= x² + y² + z²

(hopefully the way I explained the diagonals makes sense)

+ what doesn't make sense about my part 2.? (apart from it not being simplified) I haven't really done this type of question before

we have x₁ = 3x/5
so x = 5/3 *x₁, then sub that into y₁

 

[shaon0]

you've never used pythag in 3-d?
say you have 3 lines; x, y and z and you want to find the diagonal xyz
then for the xy diagonal (xy) xy² = x² + y²
and for the the xyz diagonal (xyz) xyz² = xy² + z²
= x² + y² + z²

(hopefully the way I explained the diagonals makes sense)

+ what doesn't make sense about my part 2.? (apart from it not being simplified) I haven't really done this type of question before

we have x₁ = 3x/5
so x = 5/3 *x₁, then sub that into y₁

Ok part 2 is right. My mistake sorry.
Thanks for your help

 

[lolokay]

kk, thought it was.

btw, an easy way to solve this type of locus would be to notice that the locus will have the same gradient but the y intercept will be K times as much as the old intercept, where K is the ratio of the line intersected on the origin-side

eg. for "2x-3y+6=0 and the point Q divides OP in the ratio 3:2"
you have 2x-3y+18/5=0 (since 3/5 of the line lies towards the origin, 6*3/5 = 18/5)

of course, you can't do this since in these questions because of how it asked you to solve them

 

[shaon0]

kk, thought it was.

btw, an easy way to solve this type of locus would be to notice that the locus will have the same gradient but the y intercept will be K times as much as the old intercept, where K is the ratio of the line intersected on the origin-side

eg. for "2x-3y+6=0 and the point Q divides OP in the ratio 3:2"
you have 2x-3y+18/5=0 (since 3/5 of the line lies towards the origin, 6*3/5 = 18/5)

of course, you can't do this since in these questions because of how it asked you to solve them

Thanks, for the tip.