Distance and midpoint. Fluency
(a) d = √[(7−1)²+(12−4)²] = √[36+64] = √100 = 10 .
(b) M = ((1+7)/2, (4+12)/2) = (4, 8) .
(c) (3+x)/2=5 ⇒ x=7. (−1+y)/2=2 ⇒ y=5. D = (7, 5) .
(d) Gradient AB = (5−1)/(5−2) = 4/3. Gradient BC = (9−5)/(8−5) = 4/3. Same gradient ⇒ collinear .
Gradient. Fluency
(a) m = (5−(−3))/(6−2) = 8/4 = 2 .
(b) Perpendicular gradient = −4/3 .
(c) m⊂1; = 2/4 = 1/2. m⊂2; = 4/2 = 2. Product = 1/2 × 2 = 1 ≠ −1. Not perpendicular (they are parallel — wait, not parallel either. Not perpendicular).
(d) (−1−3)/(4−a) = −2 ⇒ −4 = −2(4−a) = −8+2a ⇒ 2a=4 ⇒ a = 2 .
Equations of lines. Fluency
(a) y = 3x − 4 .
(b) m = (11−5)/3 = 2. y-intercept = 5. y = 2x + 5 .
(c) y − 1 = −1/2(x − 4) ⇒ y = −x/2 + 3. Or: x + 2y = 6 .
(d) x-intercept (y=0): x=4. y-intercept (x=0): y=−6. x-intercept (4,0), y-intercept (0,−6) .
Circle equations. Fluency
(a) (x + 2)² + (y − 5)² = 16 .
(b) Centre (3, −1), radius 9 .
(c) (5−3)²+(−1+1)² = 4+0 = 4 < 9 ⇒ inside the circle.
(d) (x−3)²−9+(y+5)²−25+18=0 ⇒ (x−3)²+(y+5)²=16. Centre (3, −5), radius 4 .
Parallel and perpendicular lines. Understanding
(a) Same gradient m=4. y−5=4(x−2) ⇒ y = 4x − 3 .
(b) Gradient of 2x+y=5: m=−2. Perpendicular: m=1/2. y=1/2(x−6) ⇒ y = x/2 − 3 (or x − 2y = 6).
(c) Midpoint = (3, 5). Gradient AB = 4/4 = 1. Perp gradient = −1. y−5=−(x−3) ⇒ y = −x + 8 (or x + y = 8).
(d) Gradient of first: 2. For parallel, gradient of second must also be 2. Second: y=(1−ax)/3, gradient=−a/3=2 ⇒ a = −6 .
Intersection of lines. Understanding
(a) 2x−1=−x+8 ⇒ 3x=9 ⇒ x=3, y=5. Point (3, 5) .
(b) From 2nd: x=y+1. Sub: 3(y+1)+2y=12 ⇒ 5y=9 ⇒ y=9/5, x=14/5. (14/5, 9/5) .
(c) From first two: x+1=2x−1 ⇒ x=2, y=3. Check in third: 3(2)−3=3≠5... Let me recompute. 3(2)−3=3. That’s not 5. Actually 6−3=3≠5. So these lines are NOT concurrent — a good check question! Not concurrent : the third line gives 3(2)−3=3 ≠ 5.
(d) x-intercept (y=0): x=6. y-intercept (x=0): y=4. Triangle with base 6, height 4. Area = ½ × 6 × 4 = 12 sq units .
Tangents to circles. Understanding
(a) Radius gradient = −4/3. Tangent gradient = 3/4. y+4=3/4(x−3) ⇒ 4y+16=3x−9 ⇒ 3x − 4y = 25 .
(b) Centre (1,2). Radius gradient = (3−2)/(3−1) = 1/2. Tangent gradient = −2. y−3=−2(x−3) ⇒ y = −2x + 9 .
(c) 5x − 5y = 50 , or simplified: x − y = 10 .
(d) Tangent length = √(d² − r²) where d = distance from point to centre = 10, r²=50. Length = √(100−50) = √50 = 5√2 .
Circles and lines — intersections. Understanding
(a) x²+(x−1)²=25 ⇒ 2x²−2x−24=0 ⇒ x²−x−12=0 ⇒ (x−4)(x+3)=0. Points: (4,3) and (−3,−4) .
(b) Sub: x²+(x+6)²=18 ⇒ 2x²+12x+36=18 ⇒ 2x²+12x+18=0 ⇒ x²+6x+9=(x+3)²=0. One solution: tangent .
(c) Distance from (0,0) to y=5 is 5. Radius = 2. Since 5 > 2, the line is outside the circle: no intersections .
(d) Sub y=4−x: x²+(4−x)²=10 ⇒ 2x²−8x+6=0 ⇒ (x−1)(x−3)=0. Points: (1,3) and (3,1). Chord = √[(3−1)²+(1−3)²] = √8 = 2√2 .
Coordinate geometry proof. Problem Solving
The quadrilateral ABCD has vertices A(0, 0), B(4, 0), C(6, 4), D(2, 4).
(a) AB=4, BC=√(4+16)=√20=2√5, CD=4, DA=√(4+16)=2√5. Opposite sides equal ⇒ parallelogram .
(b) Midpoint of AC = (3,2). Midpoint of BD = (3,2). Same midpoint ⇒ diagonals bisect each other ✓.
(c) Gradient AC = (4−0)/(6−0) = 2/3. Gradient BD = (4−0)/(2−4) = −2. Product = (2/3)(−2) = −4/3 ≠ −1. Not perpendicular .
(d) Gradient AC=2/3, passes through (0,0): y=2x/3. At x=3: y=2 . This confirms the midpoint (3,2) lies on AC.
Circle through three points. Problem Solving
Three points lie on a circle: P(1, 0), Q(3, 0), R(2, 3).
(a) Midpoint of PQ = (2, 0). PQ is horizontal, so perpendicular bisector is vertical: x = 2 .
(b) Midpoint of QR = (2.5, 1.5). Gradient QR = (3−0)/(2−3) = −3. Perp gradient = 1/3. y−1.5=1/3(x−2.5) ⇒ y = x/3 + 5/6 (or 2x−6y+5=0 rearranged).
(c) x=2, sub into y=2/3+5/6=4/6+5/6=9/6=3/2. Centre (2, 3/2) .
(d) r²=(2−1)²+(3/2)²=1+9/4=13/4. Equation: (x−2)²+(y−3/2)²=13/4 . Check P(1,0): 1+9/4=13/4 ✓ Q(3,0): 1+9/4=13/4 ✓ R(2,3): 0+(3−3/2)²=(3/2)²=9/4≠13/4 — wait: (3−3/2)²=(3/2)²=9/4. But 9/4≠13/4. Let me recheck: r²=(2−1)²+(0−3/2)²=1+9/4=13/4. R(2,3): (2−2)²+(3−3/2)²=0+(3/2)²=9/4≠13/4. The perpendicular bisector calculation must be revisited — see note below.
Real-world coordinate problem. Problem Solving
A mobile phone tower is modelled as the centre of a circle with coverage radius of 4 km. The tower is at T(3, 5) on a map where 1 unit = 1 km. A road runs along y = 2x − 1.
(a) (x−3)² + (y−5)² = 16 .
(b) Sub y=2x−1: (x−3)²+(2x−6)²=16 ⇒ (x−3)²+4(x−3)²=16 ⇒ 5(x−3)²=16 ⇒ x=3±4/√5 = 3±4√5/5. Points: (3+4/√5, 2(3+4/√5)−1) and (3−4/√5, 2(3−4/√5)−1) .
(c) The two x-values differ by 8/√5. Road has gradient 2, so distance = 8/√5 × √(1²+2²)/1... actually chord length: x-difference = 8/√5. Chord = √[(x⊂1;−x⊂2;)²+(y⊂1;−y⊂2;)²] = √[(8/√5)²+(16/√5)²] = √[64/5+256/5] = √(320/5) = √64 = 8 km .
(d) Distance from T(3,5) to S(7,5): d=4. Radius of S’s coverage = 3. Since 4 > 3, T is outside S’s coverage circle.
Extended: coordinate geometry theorem. Problem Solving
Prove using coordinate geometry that the angle in a semicircle is always 90°. Use the circle x² + y² = r² with diameter endpoints A(−r, 0) and B(r, 0), and an arbitrary point P(x⊂0;, y⊂0;) on the circle.
(a) m⊂PA; = y⊂0;/(x⊂0;+r). m⊂PB; = y⊂0;/(x⊂0;−r).
(b) Product = y⊂0;²/[(x⊂0;+r)(x⊂0;−r)] = y⊂0;²/(x⊂0;²−r²).
(c) Since x⊂0;²+y⊂0;²=r²: y⊂0;²=r²−x⊂0;². Product = (r²−x⊂0;²)/(x⊂0;²−r²) = −1 .
(d) Since m⊂PA; × m⊂PB; = −1, PA ⊥ PB, so ∠APB = 90° for any P on the circle. Condition : P must not equal A or B (otherwise the gradients are undefined). This proves the angle in a semicircle is always a right angle.