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Co-Ordinate Geometry

Co-Ordinate Geometry

Gradient, midpoint and length

Formulas Paper 1 + 2 6 min read

On this page
  1. In this note
  2. Formula bank
  3. Gradient
  4. Midpoint and length
  5. Ways they test it
  6. Mistakes to avoid
  7. Summary
  8. Practice
  9. Can I do this?
  10. Quiz

In this note

Syllabus 3.1-3.4: gradient, midpoint and length
  • 3.3Find the gradient of a line through two points.
  • 3.4Find the midpoint and the length of a line segment .
  • 3.4Work backwards: a missing end point or a missing coordinate.

Before you start: Pythagoras (for lengths). Next: The equation of a line.

Formula bank

GradientLearn it\(m=\dfrac{y_2-y_1}{x_2-x_1}\)change in y ÷ change in x; subtract in the same order top and bottom.Use it when two points given.
MidpointLearn it\(\left(\dfrac{x_1+x_2}2,\ \dfrac{y_1+y_2}2\right)\)the average of the x's and of the y's.Use it when 'midpoint'.
LengthLearn it\(\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\)Pythagoras on the change in x and change in y.Use it when 'length' or 'distance'.

Gradient

The gradient measures steepness: how far the line goes UP for each 1 it goes ACROSS. Uphill (left to right) is positive, downhill negative, horizontal 0.

General methodHow to find the gradient between two points

  1. Label the points \((x_1,y_1)\) and \((x_2,y_2)\).
  2. Change in y: \(y_2-y_1\). Change in x: \(x_2-x_1\) (same order!).
  3. Divide; simplify the fraction. Check the sign makes sense.

Example Find the gradient of the line through \((-1,\ 5)\) and \((3,\ -3)\).

Change in y over change in x.

  1. \(\dfrac{-3-5}{3-(-1)}=\dfrac{-8}{4}\)

Answer −2

Try it yourself

1
Gradient from (0, 0) to (4, 2)?
Show answer

Answer 1/2

2
Gradient from (2, 3) to (8, 11)?
Show answer

Answer 4/3

Midpoint and length

A, B and the midpoint246824681012xyA(2, 3)B(8, 11)M(5, 7)
Across 6, up 8: AB = √(36 + 64) = 10.

The midpoint is the average: \(\left(\tfrac{2+8}2,\tfrac{3+11}2\right)=(5,\ 7)\).

The length is the hypotenuse of a right-angled triangle with sides 6 and 8.

General methodHow to find the midpoint and the length

  1. Midpoint: add the x's and halve; add the y's and halve.
  2. Length: find the change in x and the change in y, square both, add, square-root.
  3. Missing end: other end = 2 × midpoint − known end (for x and y separately).

Example The midpoint of AB is (4, −1) and A is (1, 3). Find B.

B = 2 × midpoint − A.

  1. \(x_B=2(4)-1=7\)
  2. \(y_B=2(-1)-3=-5\)

Answer B(7, −5)

Try it yourself

1
Midpoint of (−2, 4) and (6, 0)?
Show answer

Answer (2, 2)

2
Length from (0, 0) to (5, 12)?
Show answer

Answer 13

Ways they test it

Type 1 Gradient, midpoint, length

"A is (−3, 1) and B is (5, −5). Find the gradient, midpoint and length of AB." (5 marks)

Spot it two points and three things to find.

  1. Gradient: Δy ÷ Δx.
  2. Midpoint: averages.
  3. Length: Pythagoras.

For example Δx = 8, Δy = −6

  1. m \(=-\tfrac34\)
  2. M(1, −2)
  3. \(\sqrt{64+36}=10\)

Answer \(-\tfrac34\), (1, −2), 10

Type 2 Missing end point

"The midpoint of AB is (4, −1); A is (1, 3). Find B." (2 marks)

Spot it a midpoint and one end given.

  1. 2 × midpoint − known end.

Type 3 Missing coordinate

"The line through (1, 3) and (4, k) has gradient 2. Find k." (2 marks)

Spot it a letter in a point.

  1. Write the gradient formula with k.
  2. Solve.

For example \(\dfrac{k-3}{4-1}=2\)

  1. \(k-3=6\)

Answer \(k=9\)

Mistakes to avoid

  • ✗ \(\dfrac{x_2-x_1}{y_2-y_1}\)✓ \(\dfrac{y_2-y_1}{x_2-x_1}\)Up over across.
  • ✗ \(\dfrac{-3-5}{-1-3}\) (orders mixed)✓ \(\dfrac{-3-5}{3-(-1)}\)Same order on top and bottom.
  • ✗ Midpoint = difference ÷ 2✓ sum ÷ 2It is an average.

Summary

  • Gradient \(=\dfrac{y_2-y_1}{x_2-x_1}\): up over across, same order.
  • Midpoint = average of the x's and of the y's.
  • Length = Pythagoras on Δx and Δy.
  • Missing end = 2 × midpoint − known end.

If you see …

  • "length of AB"Δx, Δy, then square, add, root.
  • a letter in a coordinateput it in the formula and solve.

Practice

Practice: easy, medium, hard

Easy

1
Gradient from (2, −1) to (6, 7)?
Show answer

Answer 2

2
Midpoint of (2, 3) and (8, 11)?
Show answer

Answer (5, 7)

3
Length from (2, 3) to (8, 11)?
Show answer

Answer 10

Medium

1
A(−3, 1), B(5, −5). Find the gradient of AB.
Guide me

What is it asking? a gradient

What do we know? Δy = −6, Δx = 8

Plan divide

Show working
  1. \(\dfrac{-5-1}{5-(-3)}=\dfrac{-6}8\)

Answer −3/4

2
The midpoint of AB is (4, −1) and A is (1, 3). Find B.
Guide me

What is it asking? a point

What do we know? 2 × midpoint − A

Plan x and y separately

Show working
  1. \(2(4)-1=7\)
  2. \(2(-1)-3=-5\)

Answer (7, −5)

3
Length from (−1, 5) to (3, −3), to 2 d.p.?
Guide me

What is it asking? a length

What do we know? Δx = 4, Δy = −8

Plan \(\sqrt{16+64}\)

Show working
  1. \(\sqrt{80}=8.944\ldots\)

Answer 8.94

Hard

1
The line through (1, 3) and (4, k) has gradient 2. Find k.
Guide me

What is it asking? k

What do we know? \(\tfrac{k-3}{3}=2\)

Plan solve

Show working
  1. \(k-3=6\)
  2. \(k=9\)

Answer 9

PartAnswer / workingMarksGuidance
\(\tfrac{k-3}{3}=2\)M1
9A1
Total: 2 marks
2
A(−3, 1), B(5, −5). Find the length of AB.
Guide me

What is it asking? a length

What do we know? Δx = 8, Δy = −6

Plan \(\sqrt{64+36}\)

Show working
  1. \(\sqrt{100}=10\)

Answer 10

PartAnswer / workingMarksGuidance
\(\sqrt{8^{2}+6^{2}}\)M1
10A1
Total: 2 marks
3
A(2, −1), B(6, 7). Find the length of AB to 1 d.p.
Guide me

What is it asking? a length

What do we know? Δx = 4, Δy = 8

Plan \(\sqrt{16+64}\)

Show working
  1. \(\sqrt{80}=8.94\ldots\)

Answer 8.9

PartAnswer / workingMarksGuidance
\(\sqrt{80}\)M1
8.9A1
Total: 2 marks

Can I do this?

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

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