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Coordinate Geometry IGCSE Extended

Grade 10 · Pure Mathematics · Cambridge IGCSE 0580 · Age 14–16

Welcome to Coordinate Geometry!

Coordinate geometry (also called analytic geometry) connects algebra and geometry by placing shapes on a coordinate grid. You can calculate exact distances, find midpoints, determine gradients, and work with equations of lines and circles — all using algebraic methods. This is a cornerstone topic for Cambridge IGCSE 0580 Extended and underpins A-Level mathematics.

Distance: d = √((x₂−x₁)²+(y₂−y₁)²)  |  Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2)  |  Circle: (x−a)²+(y−b)²=r²

Learning Objectives

  • Use the distance formula to find the length of a line segment
  • Find the midpoint of a line segment using the midpoint formula
  • Calculate gradients and recognise parallel and perpendicular lines
  • Find the equation of a line through two given points
  • Construct the perpendicular bisector of a line segment algebraically
  • Write and use the equation of a circle (x−a)²+(y−b)²=r²
  • Find intersections of lines with lines and curves (including circles)

Distance Formula

Length of a segment, Pythagoras in disguise

Midpoint Formula

Centre of diameter, midpoint of bisector

Gradients & Lines

Parallel, perpendicular, equation of a line

Perpendicular Bisector

Midpoint + perpendicular gradient method

Circle Equations

(x−a)²+(y−b)²=r², centre and radius

Intersections

Simultaneous equations, line meets circle

Learn 1 — Distance & Midpoint Formulas

The Distance Formula

The distance between two points A(x₁, y₁) and B(x₂, y₂) is found using Pythagoras' theorem applied to the horizontal and vertical separations.

d = √((x₂ − x₁)² + (y₂ − y₁)²)

Think of it as: draw a right-angled triangle with the line segment as the hypotenuse. The horizontal leg has length |x₂ − x₁| and the vertical leg has length |y₂ − y₁|. Pythagoras gives the hypotenuse.

Example: Find the distance between A(1, 2) and B(7, 10).
Step 1 — Identify: x₁=1, y₁=2, x₂=7, y₂=10
Step 2 — Substitute: d = √((7−1)² + (10−2)²) = √(36 + 64) = √100 = 10
Example: Find the length of segment PQ where P(−3, 4) and Q(5, −2).
d = √((5−(−3))² + (−2−4)²) = √(8² + (−6)²) = √(64 + 36) = √100 = 10
The order you subtract x₁ from x₂ does not matter because you square the result. (x₂ − x₁)² = (x₁ − x₂)². Always square before adding.

The Midpoint Formula

The midpoint M of the segment joining A(x₁, y₁) and B(x₂, y₂) is simply the average of each coordinate pair.

M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
Example: Find the midpoint of A(2, 5) and B(8, 1).
M = ((2+8)/2, (5+1)/2) = (10/2, 6/2) = (5, 3)
Example: M is the midpoint of AB. A = (−4, 3) and M = (1, 7). Find B.
Let B = (x, y).
(−4 + x)/2 = 1 → −4 + x = 2 → x = 6
(3 + y)/2 = 7 → 3 + y = 14 → y = 11
B = (6, 11)
Common error: finding B when given A and M — remember to multiply the midpoint coordinate by 2 and subtract the known endpoint, don't just double M.

Applications: Diameter and Circle Centre

If AB is a diameter of a circle, then the centre is the midpoint of AB, and the radius equals half the length of AB.

Example: A(1, 3) and B(7, 11) are endpoints of a diameter. Find the centre and radius.
Centre = midpoint = ((1+7)/2, (3+11)/2) = (4, 7)
Diameter = √((7−1)² + (11−3)²) = √(36+64) = 10
Radius = 10 ÷ 2 = 5
For any problem involving a circle defined by its diameter endpoints: (1) midpoint → centre, (2) distance formula ÷ 2 → radius. These two tools together solve the problem completely.

Learn 2 — Gradients, Parallel & Perpendicular Lines

Gradient Recap

The gradient (slope) of a line through A(x₁, y₁) and B(x₂, y₂) measures how steeply it rises or falls.

m = (y₂ − y₁) / (x₂ − x₁)

A positive gradient slopes upward left to right; a negative gradient slopes downward. A horizontal line has gradient 0; a vertical line has undefined gradient.

Example: Find the gradient of the line through (2, 3) and (6, 11).
m = (11 − 3) / (6 − 2) = 8 / 4 = 2
Example: Find the gradient through (−1, 5) and (4, −5).
m = (−5 − 5) / (4 − (−1)) = −10 / 5 = −2

Parallel Lines

Two lines are parallel if and only if they have equal gradients. They never intersect (unless they are the same line).

Parallel: m₁ = m₂
Example: Are the lines through A(0,1) & B(4,9) and through C(1,3) & D(3,7) parallel?
m₁ = (9−1)/(4−0) = 8/4 = 2
m₂ = (7−3)/(3−1) = 4/2 = 2
m₁ = m₂ = 2, so the lines are parallel.

Perpendicular Lines

Two lines are perpendicular (meet at a right angle) if the product of their gradients equals −1. Equivalently, the gradient of the perpendicular is the negative reciprocal.

Perpendicular: m₁ × m₂ = −1   ⟹   m₂ = −1/m₁
Example: A line has gradient 3. Find the gradient of a perpendicular line.
m₂ = −1/3
Example: A line has gradient −2/5. Find the perpendicular gradient.
m₂ = −1 ÷ (−2/5) = 5/2
The perpendicular gradient is NOT just the negative, and NOT just the reciprocal — it is the NEGATIVE RECIPROCAL. Flip the fraction AND change the sign.

Equation of a Line Through Two Points

Method: (1) calculate the gradient m, (2) substitute one point into y = mx + c to find c, (3) write the full equation.

Example: Find the equation of the line through (1, 5) and (3, 11).
Step 1 — Gradient: m = (11−5)/(3−1) = 6/2 = 3
Step 2 — Substitute (1,5): 5 = 3(1) + c → c = 2
Step 3 — Equation: y = 3x + 2
Example: Find the equation through (−2, 7) and (4, −5).
m = (−5−7)/(4−(−2)) = −12/6 = −2
Substitute (4,−5): −5 = −2(4) + c → c = −5+8 = 3
Equation: y = −2x + 3
Alternative: use point-slope form y − y₁ = m(x − x₁) directly. Substitute any known point and the gradient. Rearrange to y = mx + c at the end.

Learn 3 — Perpendicular Bisector, Circles & Intersections

Perpendicular Bisector of a Line Segment

The perpendicular bisector of AB is the line that: (1) passes through the midpoint of AB, and (2) is perpendicular to AB. Every point on this line is equidistant from A and B.

Method:
Step 1 — Find the midpoint M of AB.
Step 2 — Find the gradient m of AB.
Step 3 — Perpendicular gradient = −1/m.
Step 4 — Use M and the perpendicular gradient to write the line equation.
Example: Find the perpendicular bisector of A(2, 1) and B(8, 5).
Step 1 — M = ((2+8)/2, (1+5)/2) = (5, 3)
Step 2 — Gradient of AB: m = (5−1)/(8−2) = 4/6 = 2/3
Step 3 — Perpendicular gradient: −3/2
Step 4 — Line through (5,3) with gradient −3/2:
   y − 3 = −3/2 (x − 5)
   2y − 6 = −3x + 15
   2y + 3x = 21
The perpendicular bisector is used in locus problems: "Find the locus of points equidistant from A and B" is just the perpendicular bisector of AB.

Equation of a Circle

A circle with centre (a, b) and radius r has equation:

(x − a)² + (y − b)² = r²

When the centre is at the origin (0,0), this simplifies to x² + y² = r².

Example: Write the equation of a circle with centre (3, −2) and radius 5.
(x − 3)² + (y − (−2))² = 5²
(x − 3)² + (y + 2)² = 25
Example: Identify the centre and radius of (x + 4)² + (y − 1)² = 49.
Rewrite: (x − (−4))² + (y − 1)² = 7²
Centre = (−4, 1), radius = 7
Example: A circle has diameter endpoints A(1, 3) and B(7, 11). Write its equation.
Centre = midpoint = (4, 7)
Radius = (distance AB)/2 = 10/2 = 5
Equation: (x − 4)² + (y − 7)² = 25
Watch the sign: centre (3, −2) gives (x − 3)² + (y + 2)² = r². A negative y-coordinate leads to a + sign in the equation, not a minus sign.

Finding Intersections of Lines and Curves

To find where a line meets another line or curve, set the equations equal and solve simultaneously.

Example: Find the intersection of y = 2x − 1 and y = −x + 8.
Set equal: 2x − 1 = −x + 8 → 3x = 9 → x = 3
y = 2(3) − 1 = 5
Intersection: (3, 5)
Example: Find where the line y = x + 1 meets the circle x² + y² = 25.
Substitute y = x + 1 into x² + y² = 25:
x² + (x+1)² = 25
x² + x² + 2x + 1 = 25
2x² + 2x − 24 = 0
x² + x − 12 = 0
(x + 4)(x − 3) = 0 → x = −4 or x = 3
y = −3 or y = 4
Points: (−4, −3) and (3, 4)
For a line meeting a circle: always substitute the line equation into the circle equation to eliminate y (or x). You will get a quadratic — factorise or use the formula. Two solutions = two intersection points; one solution (repeated root) = tangent; no real solutions = no intersection.

Example 1 — Distance Between Two Points

Q: Find the exact distance between P(−2, 3) and Q(4, 11).
M1: d = √((4−(−2))² + (11−3)²) = √(6² + 8²)
M1: = √(36 + 64) = √100
A1: d = 10

Example 2 — Midpoint and Finding an Endpoint

Q: M(4, 6) is the midpoint of AB. A = (1, 2). Find B.
M1: (1 + x)/2 = 4 → x = 7
M1: (2 + y)/2 = 6 → y = 10
A1: B = (7, 10)

Example 3 — Equation of Line Through Two Points

Q: Find the equation of the line passing through (−1, 4) and (3, −4).
M1: m = (−4−4)/(3−(−1)) = −8/4 = −2
M1: Using (3,−4): −4 = −2(3) + c → c = 2
A1: y = −2x + 2

Example 4 — Perpendicular Bisector

Q: Find the equation of the perpendicular bisector of A(0, 4) and B(6, 0).
M1: Midpoint M = (3, 2)
M1: Gradient of AB = (0−4)/(6−0) = −2/3 → perpendicular gradient = 3/2
M1: Line through (3,2) with gradient 3/2: y − 2 = 3/2(x − 3) → 2y − 4 = 3x − 9
A1: 3x − 2y = 5 (or equivalent)

Example 5 — Equation of Circle from Diameter

Q: A(−3, 2) and B(5, 8) are the ends of a diameter. Write the equation of the circle.
M1: Centre = ((−3+5)/2, (2+8)/2) = (1, 5)
M1: Radius = (1/2)√((5−(−3))² + (8−2)²) = (1/2)√(64+36) = (1/2)×10 = 5
A1: (x−1)² + (y−5)² = 25

Example 6 — Line Meeting a Circle

Q: Find the coordinates where y = 2x − 3 meets (x−1)² + (y−2)² = 20.
M1: Substitute y = 2x−3: (x−1)² + (2x−3−2)² = 20 → (x−1)² + (2x−5)² = 20
M1: x²−2x+1 + 4x²−20x+25 = 20 → 5x²−22x+6 = 0
M1: (5x−2)(x−4) = 0 → x = 2/5 or x = 4
A1: x=4: y=5 → (4,5). x=0.4: y=−2.2 → (0.4, −2.2)

Common Mistakes in Coordinate Geometry

These are the errors examiners see most frequently. Knowing why they are wrong helps you avoid losing marks.

Mistake 1 — Subtracting coordinates in wrong order for gradient

✗ Wrong: Gradient of A(2,3) to B(6,11) → m = (2−6)/(3−11) = −4/−8 = 0.5 ... wait, that looks right? Actually wrong method — mixing up x and y:
✗ Wrong: m = (x₂−x₁)/(y₂−y₁) = (6−2)/(11−3) = 4/8 = 0.5 — numerator and denominator swapped!
✓ Correct: m = (y₂−y₁)/(x₂−x₁) = (11−3)/(6−2) = 8/4 = 2

Gradient is always rise over run: change in y DIVIDED BY change in x. "y over x" — think "up over across".

Mistake 2 — Forgetting the negative reciprocal for perpendicular lines

✗ Wrong: Line has gradient 4. Perpendicular gradient = 1/4 (just the reciprocal, forgot the negative)
✗ Wrong: Perpendicular gradient = −4 (just the negative, forgot to flip)
✓ Correct: m₁ = 4 → m₂ = −1/4 (flip AND negate)  |  Check: 4 × (−1/4) = −1 ✓

Always verify: multiply the two perpendicular gradients. The product MUST equal −1. If it doesn't, you've made an error.

Mistake 3 — Errors in the distance formula (forgetting to square root, or not squaring differences)

✗ Wrong: d = (x₂−x₁) + (y₂−y₁) — just adding the differences instead of using Pythagoras
✗ Wrong: d = (x₂−x₁)² + (y₂−y₁)² — forgetting the square root at the end
✓ Correct: d = √((x₂−x₁)² + (y₂−y₁)²) — square each difference, add, then take the square root

Write out the formula before substituting. The two most common errors are (1) not squaring and (2) not taking the square root.

Mistake 4 — Wrong sign in circle equation centre

✗ Wrong: Circle (x−3)²+(y+2)²=25 has centre (3, 2)
✓ Correct: (y+2)² = (y−(−2))², so centre is (3, −2) and radius 5

The standard form is (x−a)²+(y−b)²=r². If you see a plus sign, it means the corresponding centre coordinate is negative. Always rewrite to (x−a)² form to read off the centre safely.

Mistake 5 — Using the wrong point to find c in y = mx + c

✗ Wrong: Gradient found using A, then c found using a different unlisted coordinate by mistake
✓ Correct: After finding m, substitute EITHER of the two given points (not a made-up point). Both must give the same c — use one as a check.

After you find c, substitute the OTHER point back in to verify your equation is correct. This takes 10 seconds and catches arithmetic errors.

Mistake 6 — Forgetting the perpendicular bisector passes through the MIDPOINT

✗ Wrong: Finding perpendicular gradient correctly but then using one of the original endpoints A or B to write the line equation
✓ Correct: The perpendicular bisector passes through the MIDPOINT M of AB, not through A or B themselves

Always work in this order: (1) find midpoint M, (2) find gradient of AB, (3) find perpendicular gradient, (4) use M in the equation. Never use A or B in step 4.

Key Formulas — Coordinate Geometry

FormulaNotes / When to Use
Distance: d = √((x₂−x₁)²+(y₂−y₁)²)Length of any line segment; also find radius from centre to point on circle
Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2)Centre of a diameter; midpoint for perpendicular bisector; locus equidistant from two points
Gradient: m = (y₂−y₁)/(x₂−x₁)Rise over run; steepness of a line
Parallel lines: m₁ = m₂Equal gradients → parallel (and different y-intercepts)
Perpendicular gradient: m₂ = −1/m₁Negative reciprocal; product m₁×m₂ = −1
Line equation: y = mx + cUse gradient + one point to find c
Point-slope form: y − y₁ = m(x − x₁)Direct method when you have a point and gradient
Circle: (x−a)²+(y−b)²=r²Centre (a,b), radius r; origin-centred: x²+y²=r²
Perpendicular bisector of AB: midpoint of AB + perpendicular gradient to AB
Intersection: substitute one equation into the other → solve simultaneously

Key Relationships to Remember

Diameter → Circle: Midpoint of diameter = centre; (½ × distance AB) = radius
Tangent to circle at point P: The tangent is perpendicular to the radius at P
Locus equidistant from A and B: Perpendicular bisector of AB
Number of intersections of line and circle: discriminant of quadratic → 0 (tangent), >0 (2 points), <0 (none)

Coordinate Geometry Explorer

Enter two points A and B. The tool will calculate the distance, midpoint, gradient, line equation, and draw everything on the coordinate grid.

Enter two points and click "Calculate & Draw".

Exercise 1 — Distance Formula

1. Find the distance between A(0, 0) and B(3, 4).

2. Find the distance between P(1, 1) and Q(4, 5).

3. Find the distance between A(0, 3) and B(4, 0). Give exact answer.

4. Find the distance between C(−2, 1) and D(4, 9).

5. Find the distance between P(3, −1) and Q(−3, 7). Give exact value.

6. A circle has centre O(0,0) and passes through (5, 12). What is the radius?

7. Find the distance between A(−4, −3) and B(8, 2). Give answer to 2 d.p.

8. The distance between (k, 0) and (0, 8) is 10. Find the positive value of k.

Exercise 2 — Midpoint Formula

1. Find the midpoint of A(2, 4) and B(8, 10). Enter x then y.

2. Find the midpoint of P(−3, 5) and Q(7, 1). Enter x then y.

3. M(5, 3) is the midpoint of AB. A = (2, 1). Find B. Enter x then y.

4. M(0, 4) is the midpoint of PQ. P = (−4, 7). Find Q. Enter x then y.

5. A(1, 3) and B(7, 11) are the ends of a diameter. Find the centre. Enter x then y.

6. Find the midpoint of (−6, 0) and (0, −8). Enter x then y.

7. M(3, −1) is the midpoint of AB. A = (7, 4). Find B. Enter x then y.

8. A(−5, 2) and B(3, −6) form a diameter. Find the centre. Enter x then y.

Exercise 3 — Gradients & Equations of Lines

1. Find the gradient of the line through (1, 3) and (5, 11).

2. Find the gradient of the line through (−2, 6) and (4, −3). Give as a decimal.

3. A line has gradient 4. What is the gradient of a perpendicular line?

4. A line has gradient −3/2. What is the gradient of a perpendicular line?

5. Find the y-intercept (c) of the line through (2, 9) with gradient 3.

6. Find the y-intercept (c) of the line through (−1, 5) with gradient −2.

7. Find the gradient of the line through (0, −3) and (6, 9).

8. A line has gradient 5/3. Find the gradient of a perpendicular line as a decimal.

Exercise 4 — Perpendicular Bisector & Circle Equations

1. Circle (x−2)²+(y−5)²=36. What is the radius?

2. Circle (x+3)²+(y−1)²=16. What is the x-coordinate of the centre?

3. Write the equation of the circle with centre (0,0) and radius 7. Enter r² (the right-hand side).

4. A circle has centre (1, 4) and passes through (4, 8). Find the radius.

5. Diameter endpoints: A(2, 0) and B(8, 6). Find the radius of the circle.

6. Perpendicular bisector of A(0, 0) and B(4, 0). What is the x-coordinate where it crosses the x-axis?

7. Find the gradient of the perpendicular bisector of A(1, 3) and B(5, 7).

8. Circle (x−3)²+(y+4)²=25. Does the point (3, 1) lie on, inside, or outside? Enter 0=on, −1=inside, 1=outside.

Exercise 5 — Intersections & Mixed

1. Lines y = 3x − 2 and y = x + 4 intersect. Find the x-coordinate.

2. Lines y = 3x − 2 and y = x + 4 intersect. Find the y-coordinate.

3. Line y = x + 5 meets x² + y² = 25. Find the larger x-coordinate of intersection.

4. Line y = x + 5 meets x² + y² = 25. Find the smaller x-coordinate of intersection.

5. Do lines y = 2x+1 and 2y = 4x+5 intersect? Enter 1 for yes, 0 for no (parallel).

6. The line y = k is tangent to the circle x² + y² = 25. Find the positive value of k.

7. Find the x-coordinate where y = −x + 3 meets y = x − 1.

8. How many intersections does x² + y² = 9 have with the line y = x + 5? (0, 1, or 2)

Practice — 25 Mixed Questions

Non-calc = exact answers. Calc = decimals acceptable.

1. Distance between (0,0) and (5,12).

2. Midpoint of (−4, 6) and (10, −2). Enter x-coordinate.

3. Midpoint of (−4, 6) and (10, −2). Enter y-coordinate.

4. Gradient through (0,1) and (4,9).

5. Perpendicular gradient to a line with gradient 2.

6. y-intercept of the line through (3,7) with gradient 2.

7. Distance between (−1,−1) and (2,3). Give exact value.

8. M(2,5) is midpoint of AB. A=(−2,3). Find the x-coordinate of B.

9. M(2,5) is midpoint of AB. A=(−2,3). Find the y-coordinate of B.

10. Circle centre (4,−3) radius 6. Write r² (right-hand side of equation).

11. Circle (x+1)²+(y−3)²=49. Find the radius.

12. Diameter endpoints A(0,0) and B(6,8). Find the radius.

13. Find gradient of perpendicular bisector of A(0,4) and B(4,0).

14. Lines y=5x−3 and y=5x+7. Are they parallel? Enter 1 for yes, 0 for no.

15. x-intercept of the line y=3x−9 (set y=0, find x).

16. Distance between (2,3) and (2,10).

17. Find the y-intercept (c) of line through (4,5) with gradient −1.

18. Circle (x−2)²+(y+5)²=100. What is the y-coordinate of the centre?

19. Gradient through (−3,−3) and (6,6).

20. Perpendicular gradient to line with gradient −1/3.

21. Find the x-coordinate of intersection of y=2x+1 and y=−x+7.

22. Find x-midpoint of (3,−5) and (7,3).

23. Find y-midpoint of (3,−5) and (7,3).

24. A circle has x²+y²=r². It passes through (8,6). Find r².

25. Diameter endpoints (−2,3) and (6,7). Find the y-coordinate of the centre.

Challenge — 12 Questions (IGCSE Extended Level)

1. A(−1, 3) and B(5, 11) are the ends of a diameter. Find the equation of the circle in the form (x−a)²+(y−b)²=r². Enter r².

2. The perpendicular bisector of A(2, 4) and B(8, 2) passes through point (5, 3). Find the y-intercept of this perpendicular bisector.

3. A line through (k, 1) and (3, 7) has gradient 2. Find k.

4. Find the distance between the intersections of y = x + 1 and x² + y² = 25. Give answer to 2 d.p.

5. A triangle has vertices A(0,0), B(6,0), C(3,4). Find the length of the median from A to the midpoint of BC.

6. Show that A(1,2), B(4,6), C(8,3), D(5,−1) form a rectangle. What is the length of diagonal AC? Give exact value.

7. The line y = mx + 2 is tangent to the circle x² + y² = 5. Find the positive value of m. (Hint: tangent means discriminant = 0)

8. A(−2,1), B(4,3), C(5,0). Prove or disprove that angle ABC = 90°. Enter the product m_AB × m_BC.

9. The circle (x−3)²+(y−1)²=25 and the line x=3 intersect at two points. Find the y-coordinate of the upper intersection point.

10. Find the equation of the line perpendicular to y=3x−5 that passes through (6,1). Enter the y-intercept c.

11. A(0,0), B(4,0), C(4,3). Find the exact distance from A to the midpoint of BC.

12. Two perpendicular lines have gradients m and (m+3). Find the two possible values of m. Enter the positive value.

Exam Style Questions

Mark-scheme style. Show all working in your book. Enter final answers for self-marking.

Question 1 — Distance, Midpoint and Circle [6 marks]

A(−2, 3) and B(6, 9) are the endpoints of a diameter of a circle.

(a) Find the coordinates of the centre of the circle. Enter x then y.
(b) Calculate the radius of the circle. Give your answer in surd form (enter the value under the square root, i.e. enter 50 if radius = √50, or 5 if r = 5).
(c) Write down the equation of the circle.
Enter r² (right-hand side):

Question 2 — Perpendicular Bisector [5 marks]

A = (1, 7) and B = (9, 3).

(a) Find the gradient of AB.
(b) Find the equation of the perpendicular bisector of AB. Enter the y-intercept c (in y = mx + c form where m = 2).
(c) The perpendicular bisector meets the y-axis at point P. Find the y-coordinate of P (this equals c from part b).

Question 3 — Equation of Line and Intersection [5 marks]

Line L passes through (2, 5) and (6, 13).

(a) Find the gradient of L.
(b) Find the equation of L in the form y = mx + c. Enter c.
(c) Line M has equation y = −x + 11. Find the x-coordinate of the intersection of L and M.

Question 4 — Circle and Line Intersection [6 marks]

A circle has equation x² + y² = 50. A line has equation y = x + 2.

(a) Show that the x-coordinates of the intersection satisfy 2x² + 4x − 46 = 0. How many solutions does this quadratic have? Enter the discriminant (b²−4ac).
(b) Solve 2x² + 4x − 46 = 0 (or equivalent x² + 2x − 23 = 0). Find the larger x-coordinate to 2 d.p.
(c) Find the corresponding y-coordinate for the larger x-value, to 2 d.p.

Question 5 — Proof and Mixed Coordinate Geometry [5 marks]

Points P(0, 4), Q(4, 6), R(8, 4) and S(4, 2) are given.

(a) Find the gradient of PQ.
(b) Find the gradient of QR.
(c) Find the distance PR (diagonal). Give exact value.
(d) What type of quadrilateral is PQRS? Identify by entering: 1=rhombus, 2=rectangle, 3=square, 4=kite