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O-Level Mathematics · Secondary 3 and 4

Coordinate Geometry

Coordinate geometry uses the x and y coordinates of points to answer questions about straight lines. In O-Level Mathematics you find the gradient of a line through two points, the length of the line segment joining them and the equation of the line, and you use these to solve problems about shapes on a grid. The topic links algebra to graphs, and the same formulas reappear in Additional Mathematics, where they are extended to perpendicular lines and circles.

What the syllabus expects

  • Find the gradient of a straight line given the coordinates of two points on it
  • Find the length of a line segment from the coordinates of its end points
  • Find the equation of a straight line in the form y = mx + c, given its gradient and a point, or two points
  • Read the gradient and y-intercept from an equation, rearranging to y = mx + c when needed
  • Use the fact that parallel lines have equal gradients
  • Solve geometric problems involving points and lines on a coordinate grid, such as areas of triangles

Gradient

The gradient of a line measures its steepness: gradient = (change in y)/(change in x) = (y₂ − y₁)/(x₂ − x₁). For C(1, 2) and D(5, −6), the gradient is (−6 − 2)/(5 − 1) = −8/4 = −2. Subtract the coordinates in the same order on the top and the bottom. A line sloping up from left to right has a positive gradient and one sloping down has a negative gradient. A horizontal line has gradient 0, because y does not change. A vertical line has no defined gradient, because the change in x is 0 and you cannot divide by zero. Lines that are parallel have the same gradient, so any line parallel to y = 3x − 4 also has gradient 3. When a line is given in a form such as 3x + 2y = 8, rearrange it to y = −(3/2)x + 4 before reading the gradient, which is −3/2.

Length of a line segment

The length of the segment between (x₁, y₁) and (x₂, y₂) comes from Pythagoras' theorem. The horizontal and vertical distances form the two shorter sides of a right-angled triangle, so length = √((x₂ − x₁)² + (y₂ − y₁)²). For C(1, 2) and D(5, −6), the length is √(4² + (−8)²) = √(16 + 64) = √80 = 8.94 units to 3 significant figures. Because the differences are squared, the order of subtraction does not matter here, and negative differences become positive. Keep an exact surd such as √80 until the end if the answer feeds into a later part or the question asks for an exact value. Length questions often ask you to show that a triangle is isosceles or right-angled. For a right angle, check whether the squares of the two shorter sides add up to the square of the longest side, which is the converse of Pythagoras' theorem.

Equation of a straight line

Every non-vertical straight line has an equation of the form y = mx + c, where m is the gradient and c is the y-intercept. To find the equation through a point with a known gradient, substitute the gradient for m and the coordinates of the point for x and y, then solve for c. For a line through (−1, 4) with gradient −2, 4 = −2(−1) + c, so c = 2 and the line is y = −2x + 2. With two points, find the gradient first and then use either point. Horizontal lines have equations of the form y = k and vertical lines have equations of the form x = k, for a constant k. To find where a line crosses the x-axis, put y = 0, and for the y-axis put x = 0. To find where two lines meet, solve their equations simultaneously.

Problems on a grid

Many questions combine these skills with geometry. To find the area of a triangle with a horizontal or vertical side, use that side as the base and read the perpendicular height from the coordinates. For the triangle with vertices P(1, 1), Q(7, 1) and R(3, 5), the base PQ is 6 units, the height is 5 − 1 = 4 units, and the area is ½ × 6 × 4 = 12 square units. If no side is horizontal or vertical, enclose the shape in a rectangle and subtract the right-angled triangles around it. To show that a quadrilateral is a parallelogram, show that both pairs of opposite sides have equal gradients. A rough sketch helps you see which sides to use and whether an answer is sensible: a gradient that comes out positive for a line you drew sloping down is a sign of a subtraction slip.

Worked example

The points A(−2, 1) and B(4, 9) are given. Find (a) the gradient of AB, (b) the length of AB, (c) the equation of the line AB, (d) the coordinates of the point where the line AB meets the x-axis.

  1. Gradient = (9 − 1)/(4 − (−2)) = 8/6 = 4/3.
  2. Length = √((4 − (−2))² + (9 − 1)²) = √(36 + 64) = √100 = 10 units.
  3. Use y = mx + c with m = 4/3 and the point B(4, 9): 9 = (4/3)(4) + c = 16/3 + c, so c = 11/3.
  4. The equation is y = (4/3)x + 11/3, which can also be written as 3y = 4x + 11.
  5. Check with A: (4/3)(−2) + 11/3 = −8/3 + 11/3 = 1, which is correct.
  6. For the x-axis, put y = 0: 0 = 4x + 11, so x = −11/4 = −2.75.

Answer: (a) 4/3 (b) 10 units (c) 3y = 4x + 11 (d) (−2.75, 0)

Common mistakes

  • Mixing the order of subtraction. (9 − 1)/(−2 − 4) gives −4/3 instead of 4/3. Use the same point first on the top and on the bottom.
  • Putting x over y. Gradient is the change in y divided by the change in x. Inverting it gives 3/4 instead of 4/3.
  • Reading the gradient from the wrong form. In 2x + 3y = 12 the gradient is not 2. Rearrange to y = −(2/3)x + 4 first, so the gradient is −2/3.
  • Losing the sign when subtracting a negative coordinate. 4 − (−2) = 6, not 2. Write the brackets in every time.
  • Confusing the equations of horizontal and vertical lines. A horizontal line through (3, 5) is y = 5, and a vertical line through it is x = 3.

Exam tips

  • Draw a quick sketch with the points roughly in place. It tells you whether the gradient should be positive or negative and whether a length is reasonable.
  • Leave gradients as fractions such as 4/3 rather than 1.33, so the equation of the line stays exact and later substitutions give exact answers.
  • Check every equation you find by substituting the coordinates of a second point that should lie on the line.

Practice questions

1. Find the gradient and the length of the line segment joining P(1, −3) and Q(5, 0).

Show answer

Gradient 3/4 and length 5 units. Gradient = (0 − (−3))/(5 − 1) = 3/4. Length = √(4² + 3²) = √25 = 5.

2. Find the equation of the line that passes through (2, 5) and is parallel to y = 3x − 4.

Show answer

y = 3x − 1. Parallel lines have equal gradients, so m = 3. Then 5 = 3(2) + c gives c = −1.

3. The line 2x + 3y = 12 meets the x-axis at A and the y-axis at B, and O is the origin. Find the gradient of AB, the length of AB and the area of triangle OAB.

Show answer

Putting y = 0 gives A(6, 0), and putting x = 0 gives B(0, 4). Gradient = (4 − 0)/(0 − 6) = −2/3. Length AB = √(6² + 4²) = √52 = 7.21 units to 3 significant figures. Area of OAB = ½ × 6 × 4 = 12 square units.

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