Gradient from Two Points

\( m = \frac{y_2 - y_1}{x_2 - x_1} \)
Coordinate Geometry GCSE

Find the gradient between (-1,-1) and (2,5).

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
Subtract y-values then x-values.

Explanation

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Statement

The gradient (or slope) of a straight line through two points shows how steep the line is. It measures the change in the vertical direction compared with the change in the horizontal direction. Given two points \((x_1, y_1)\) and \((x_2, y_2)\), the gradient is calculated using:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Here, \(m\) stands for the gradient of the line.

Why it’s true

  • The numerator \(y_2 - y_1\) represents the vertical change (also called the "rise").
  • The denominator \(x_2 - x_1\) represents the horizontal change (also called the "run").
  • Dividing rise by run gives a consistent measure of steepness that works for any two distinct points on a straight line.

Recipe (how to use it)

  1. Identify the coordinates of the two points. Label them clearly as \((x_1, y_1)\) and \((x_2, y_2)\).
  2. Subtract the y-coordinates: compute \(y_2 - y_1\).
  3. Subtract the x-coordinates: compute \(x_2 - x_1\).
  4. Divide the vertical change by the horizontal change: \(\frac{y_2 - y_1}{x_2 - x_1}\).
  5. Simplify the fraction or convert to a decimal if appropriate.

Spotting it

You are expected to use this formula when you are given two points and asked about the slope of the line between them. Common contexts include:

  • Analysing linear graphs in coordinate geometry.
  • Determining if two lines are parallel (equal gradients) or perpendicular (gradients multiply to \(-1\)).
  • Forming the equation of a straight line when combined with the point–slope form.

Common pairings

  • Equation of a line: \(y - y_1 = m(x - x_1)\).
  • Perpendicular gradients: \(m_1 \cdot m_2 = -1\).
  • Parallel lines: equal gradients.

Mini examples

  1. Given: Points (1, 2) and (3, 6). Find: Gradient. Answer: \(\frac{6 - 2}{3 - 1} = \frac{4}{2} = 2\).
  2. Given: Points (0, 0) and (4, -2). Find: Gradient. Answer: \(\frac{-2 - 0}{4 - 0} = \frac{-2}{4} = -\tfrac{1}{2}\).

Pitfalls

  • Forgetting order: Always subtract in the same order for both x and y. Mixing them gives the wrong sign.
  • Dividing the wrong way round: Remember gradient is rise over run, not run over rise.
  • Zero denominator: If \(x_1 = x_2\), the line is vertical and the gradient is undefined.
  • Sign mistakes: Pay attention to negatives when subtracting.

Exam strategy

  • Write the formula first before substituting numbers. This helps avoid mistakes.
  • Show working clearly: subtract y-values and x-values on separate lines.
  • Check if your gradient is positive or negative by looking at the graph: if the line slopes upwards, gradient should be positive.

Summary

The gradient formula \(\frac{y_2 - y_1}{x_2 - x_1}\) is a central tool in coordinate geometry. It measures steepness consistently and connects with many other ideas such as parallel and perpendicular lines, linear equations, and proportional reasoning. Mastery of this formula gives confidence in analysing and constructing straight-line graphs.

Worked examples

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  1. Find the gradient between (1,2) and (5,6).
    1. \( y2 - y1 = 6 - 2 = 4 \)
    2. \( x2 - x1 = 5 - 1 = 4 \)
    3. \( Gradient = 4/4 = 1 \)
    Answer: 1
  2. Calculate the gradient between (-2,3) and (4,9).
    1. \( y2 - y1 = 9 - 3 = 6 \)
    2. \( x2 - x1 = 4 - (-2) = 6 \)
    3. \( Gradient = 6/6 = 1 \)
    Answer: 1
  3. Find the gradient between (0,0) and (4,-2).
    1. \( y2 - y1 = -2 - 0 = -2 \)
    2. \( x2 - x1 = 4 - 0 = 4 \)
    3. \( Gradient = -2/4 = -1/2 \)
    Answer: -\tfrac{1{2
  4. Find the gradient of the line through (3,7) and (6,7).
    1. \( y2 - y1 = 7 - 7 = 0 \)
    2. \( x2 - x1 = 6 - 3 = 3 \)
    3. \( Gradient = 0/3 = 0 \)
    Answer: 0
  5. Find the gradient between (2,-3) and (2,5).
    1. \( y2 - y1 = 5 - (-3) = 8 \)
    2. \( x2 - x1 = 2 - 2 = 0 \)
    3. Division by zero → undefined gradient
    Answer: undefined
  6. Find the gradient between (-4,-2) and (2,1).
    1. \( y2 - y1 = 1 - (-2) = 3 \)
    2. \( x2 - x1 = 2 - (-4) = 6 \)
    3. \( Gradient = 3/6 = 1/2 \)
    Answer: \tfrac{1{2
  7. Find the gradient of the line through (10,15) and (13,3).
    1. \( y2 - y1 = 3 - 15 = -12 \)
    2. \( x2 - x1 = 13 - 10 = 3 \)
    3. \( Gradient = -12/3 = -4 \)
    Answer: -4
  8. Calculate the gradient between (1,8) and (7,2).
    1. \( y2 - y1 = 2 - 8 = -6 \)
    2. \( x2 - x1 = 7 - 1 = 6 \)
    3. \( Gradient = -6/6 = -1 \)
    Answer: -1
  9. Calculate the gradient between (-5,-1) and (0,-6).
    1. \( y2 - y1 = -6 - (-1) = -5 \)
    2. \( x2 - x1 = 0 - (-5) = 5 \)
    3. \( Gradient = -5/5 = -1 \)
    Answer: -1
  10. Find the gradient between (-3,4) and (5,-2).
    1. \( y2 - y1 = -2 - 4 = -6 \)
    2. \( x2 - x1 = 5 - (-3) = 8 \)
    3. \( Gradient = -6/8 = -3/4 \)
    Answer: -\tfrac{3{4