(i) (2,3) and (4,1)

d = √[(4-2)² + (1-3)²] = √[2² + (-2)²] = √[4 + 4] = √8 = 2√2 units

(ii) (-5,7) and (-1,3)

d = √[(-1+5)² + (3-7)²] = √[4² + (-4)²] = √[16 + 16] = √32 = 4√2 units

(iii) (a,b) and (-a,-b)

d = √[(-a-a)² + (-b-b)²] = √[(-2a)² + (-2b)²] = √[4a² + 4b²] = 2√(a²+b²) units

d = √[(36-0)² + (15-0)²] = √[1296 + 225] = √1521 = 39 units

Area = ½|1(3+11) + 2(-11-5) + (-2)(5-3)|

= ½|1×14 + 2×(-16) + (-2)×2| = ½|14 - 32 - 4| = ½×22 = 11

Area ≠ 0, so not collinear.

AB = √[(6-5)² + (4+2)²] = √[1+36] = √37

BC = √[(7-6)² + (-2-4)²] = √[1+36] = √37

AC = √[(7-5)² + (-2+2)²] = √[4+0] = 2

AB = BC, so yes, isosceles triangle.

AB = √[(1+1)²+(0+2)²] = √[4+4] = 2√2

BC = √[(-1-1)²+(2-0)²] = √[4+4] = 2√2

CD = √[(-3+1)²+(0-2)²] = √[4+4] = 2√2

DA = √[(-1+3)²+(-2-0)²] = √[4+4] = 2√2

All sides equal. Diagonals: AC = 4, BD = 4 → Square

This is a general quadrilateral (no special properties).

AB = √10, BC = √18, CD = √10, DA = √18 (opposite sides equal)

Diagonals: AC = 2, BD = √52 → not equal → Parallelogram

Let point be (x,0). Then (x-2)² + 25 = (x+2)² + 81

x² - 4x + 29 = x² + 4x + 85 → -8x = 56 → x = -7

Point = (-7, 0)

√[(10-2)² + (y+3)²] = 10 → 64 + (y+3)² = 100

(y+3)² = 36 → y+3 = ±6 → y = 3 or y = -9

QP² = 41, QR² = x² + 25 → x² + 25 = 41 → x² = 16 → x = ±4

QR = √(16+25)=√41, PR when x=4: √[(4-5)²+(6+3)²] = √[1+81]=√82

(x-3)²+(y-6)² = (x+3)²+(y-4)²

x²-6x+9+y²-12y+36 = x²+6x+9+y²-8y+16

-6x-12y+45 = 6x-8y+25 → -12x-4y = -20 → 3x + y = 5

x = (2×4 + 3×-1)/5 = (8-3)/5 = 1

y = (2×-3 + 3×7)/5 = (-6+21)/5 = 3

Point = (1, 3)

1st (1:2): x = (1×-2+2×4)/3 = 2, y = (1×-3+2×-1)/3 = -5/3

2nd (2:1): x = (2×-2+1×4)/3 = 0, y = (2×-3+1×-1)/3 = -7/3

Points: (2, -5/3) and (0, -7/3)

Let ratio = k:1. Then (6k-3)/(k+1) = -1 → 6k-3 = -k-1 → 7k=2 → k=2/7

Ratio = 2:7

Let ratio = k:1. On x-axis y=0 → (5k-5)/(k+1)=0 → k=1

Point: x = (1×-4+1×1)/2 = -3/2 → Ratio 1:1, point (-3/2,0)

Centre is midpoint: (x+1)/2=2 → x=3, (y+4)/2=-3 → y=-10

A(3, -10)

AP:PB = 3:4 → x = (3×2+4×-2)/7 = (6-8)/7 = -2/7

y = (3×-4+4×-2)/7 = (-12-8)/7 = -20/7

P(-2/7, -20/7)

1st point (1:3): x = (1×2+3×-2)/4 = -1, y = (1×8+3×2)/4 = 3.5

2nd point (2:2): x = 0, y = 5

3rd point (3:1): x = (3×2+1×-2)/4 = 1, y = (3×8+1×2)/4 = 6.5

Diagonal AC = √[(-1-3)²+(4-0)²] = √[16+16] = 4√2

Diagonal BD = √[(-2-4)²+(-1-5)²] = √[36+36] = 6√2

Area = ½ × d₁ × d₂ = ½ × 4√2 × 6√2 = ½ × 24 × 2 = 24 sq units

(i) Area = ½|2(0+4)+(-1)(-4-3)+2(3-0)| = ½|8+7+6| = 10.5 sq units

(ii) Area = ½|(-5)(-5-2)+3(2+1)+5(-1+5)| = ½|35+9+20| = 32 sq units

ΔABC = ½|1(2-3)+6(3-2)+5(2-2)| = 2.5

ΔADC = ½|1(3-4)+5(4-2)+3(2-3)| = 3

Total = 5.5 sq units

Let point be (x,0). (x-2)²+25 = (x+2)²+81 → -8x=56 → x=-7

Point = (-7, 0)

Let point be (0,y). 1+(2-y)² = 16+(3-y)² → 5-4y+y²=25-6y+y² → 2y=20 → y=10

Point = (0, 10)

Area = ½|(-4)(-5-3)+(-3)(3+2)+3(-2+5)+2(-2+2)| = ½|32-15+9+0| = 13 sq units

Let centre O(x,y). OA² = OB² = OC²

From OA²=OB²: (x-6)²+(y+6)² = (x-3)²+(y+7)² → -6x-2y=-14 → 3x+y=7

From OA²=OC²: (x-6)²+(y+6)² = (x-3)²+(y-3)² → -6x+18y=-54 → x-3y=9

Solving: x=3, y=-2 → Centre = (3, -2)