maths locus help

20.
(a)
Slope of L1 = -1 / (-√3) = 1 / √3
Slope of L2 = -(-√3) / 1 = √3

Their slopes are NOT equal.
Hence, L1 and L2are NOT parallel.

(b)
Let (x1, y1) the coordinates of the point.

The point is equidistant from L1 and L2.
|x1 - √3y1| / √[1² + (√3)²] = |y1 - √3x1|/ √[1² + (√3)²]
x1 - √3y1 = y1 - √3x1 ..or.. x1 - √3y1= -(y1 - √3x1)
(1 + √3)x1 - (1 + √3)y1 = 0 ..or.. (1 - √3)x1 + (1 - √3)y1 = 0
x1 - y1 = 0 ..or.. x1 + y1= 0

The equation of the locus is :
x - y = 0 ..or.. x + y = 0


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21.
(a) (i)
Let (x1, y1) be the coordinates of the point.

The point is equidistant from A and B :
√[(x1 + 5)² + (y1 - 1)²] = √[(x1 + 3)² + (y1- 5)²]
x1² + 10x1 + 25 + y1² - 2y1 + 1 = x1²+ 6x1 + 9 + y1² - 10y1 + 25
4x1 + 8y1 - 8 = 0
x1 + 2y1 - 2 = 0

The equation of the locus of the point : x+ 2y - 2 = 0

(a) (ii)
Let (x2, y2) be the coordinates of the point.

The point is equidistant from B and C :
√[(x2 + 3)² + (y2 - 5)²] = √[(x2 - 3)² + (y2+ 7)²]
x2² + 6x2 + 9 + y2² - 10y2 + 25 = x²- 6x2 + 9 + y2² + 14y2 + 49
12x2 - 24y2 - 24 = 0
x2 - 2y2 - 2 = 0

The equation of the locus of the point : x- 2y - 2 = 0

(b) (i)
x + 2y - 2 = 0 ...... [1]
x - 2y - 2 = 0 ...... [2]

[1] - [2]:
4y = 0
y = 0

Put y = 0 into [1] :
x + 2(0) - 2 = 0
x = 2

Coordinates of D = (2, 0)

(b) (ii)
Let (x3, y3) be the coordinates of P.

PD = AD
√[(x3 - 2)² + (y3 - 0)²] = √[(-5 - 2)² + (1 - 0)²]
x3² - 4x3 + 4 + y3² = 50
x3² + y3² - 4x3 - 46 = 0

Equation of the locus of P : x² + y²- 4x - 46 = 0

Q20a)
Method 1: rewrite L1 and L2 in the form of y=mx+c, where m is the slope
L1: y=x/√3 ==> slope of L1 (m1) = 1/√3
L2: y =x√3 ==> slope of L2 (m2) = √3

Method 2: for straight lines in the form ax+by+c=0, slope of the line = -a/b
m1= -(-1)/√3 = 1/√3
m2 = -(-√3)/1 = √3

since m1 doesnt equal m2, L1 and L2 are not parallel.

Q20b)
「CONCEPT:
if the included angle(夾角) between two lines with slopes m1 and m2 is A, then
tanA = (m2-m1)/(1+m1m2)」

the required locus is a straight line, so let the locus be y=kx+c
the included angle between the locus and L1 equals that between the locus and L2, with the help of the above concept:
(k-m1)/(1+km1) = (m2-k)/(1+km2)
(k-1/√3)/(1+k/√3) = (√3-k)/(1+k√3)
(k√3-1)/(√3+k) = (√3-k)/(1+k√3)
3k²-1 = 3-k²
4k²=4
k = 1 or -1
note that L1 and L2 both pass through the origin
so put (x,y)=(0,0) into the equation of locus gives c=0
so the required equation of loci are
y=x or y=-x, ie, x-y=0 or x=y=0 -----##

Q21a) the question is asking the eqt. of perpendicular bisector
i)
slope of AB = (5-1)/(-3-(-5))=2
mid-pt of AB = [(-3-5)/2, (5+1)/2] = (-4,3)
slope of the locus = -1/2
locus: y=-x/2 +c
put (-4,3) into the locus, solving gives c = 1
so the required locus: y=-x/2+1 or x+2y-2 = 0

ii)similar to i)
the required locus: y=x/2-1 or x-2y-2 = 0

b)
CONCEPT 1: circumcentre of a triangle is a point equidistant to all vertices
CONCEPT 2: circumcentre of a triangle is the intersection of the perpendicular bisectors of the three sides (or any 2 sides)

so solving the two equation of locus in part (a) gives the answer to Q21b(i)
(i)
y=-x/2+1
y=x/2-1
solving gives (x,y) = (2,0)
so the coordinates D is (2,0)

(ii)
since D is the circumcentre of triangle ABC
D is the centre of the the circle ABC
since AD=PD
locus of P is equation of a circle with AD as radius and D as centre
length of AD = sqrt( (1-0)² + (-5-2)²) = sqrt50
so the required locus : (x-2)²+(y-0)²=50
ie, x²+y²-4x-46=0

Question 20

slope of L1 = 1/sqrt(3)
slope of L2 = sqrt(3)
So they are not parallel.

L1: x-sqrt(3)y=0 and L2 : sqrt(3)x-y=0
Let P(h,k) be the point.
absolute value((h-sqrt(3)k)/sqrt(1^2+3)) = absolute value((sqrt(3)h-k)/sqrt(1^2+3))
This gives h - sqrt(3)k = sqrt(3)h - k or h - sqrt(3)k = -sqrt(3)h + k
Both equation give h - k = 0.
The required equation of locus is x - y = 0