Year 11 2u help. Please show all working (1 Viewer)

[kpad5991]

Q1. A point P(x,y) moves so that it's distance from the point K(2,5) is twice it's distance from the line x=-1. Draw a diagram and find the equation of the locus of P.

Q2. Sketch the locus of all points that are 1 unit from the square with centre the origin and vertices A(1,1), B (-1,1), C(-1,-1) and D(1,-1)

Q3. Find the locus of all points equidistant from the three points A(3,0), B(0,9) and C(7,8).

 

[Sp3ctre]

I only really have time to do one more question rn so I'll leave the other two for others to answer, might come back in a few days if no one has replied by then.

1. Distance of PK via distance formula = sqrt[(x-2)^2 + (y-5)^2]

The distance of P from x=-1, if you just think about it logically, will always be a horizontal distance since x=-1 is a vertical line. So the distance will be the absolute value of x coordinate of P - x coordinate of x=-1, i.e. |x-(-1)| = |x+1|

Now you have sqrt[(x-2)^2 + (y-5)^2] = 2|x+1| (since distance from K is twice the distance from x=-1)

Square both sides:
(x-2)^2 + (y-5)^2 = 4(x+1)^2
x^2 - 4x + 4 + y^2 - 10y + 25 = 4(x^2 + 2x + 1)
x^2 - 4x + y^2 - 10y + 29 = 4x^2 + 8x + 4
-3x^2 - 12x + y^2 - 10y + 25 = 0
3x^2 + 12 - y^2 + 10y - 25 = 0

You could probably leave that as the equation, but using 4 unit techniques you can rearrange that into the basic equation of a hyperbola. I can't really give you a diagram here but it is a hyperbola centered at (-2,5).

 

[Thatstudentm9]

using A(1,1) as the center
sub into formula (x-h)^2+(y-k)^2 =1^2
ans is as
repeat the process for all the origins hope this helped