Learning Objectives
  • Use Cartesian coordinates and draw straight-line graphs.
  • Calculate gradients, distances and midpoints from coordinates.
  • Find equations of straight lines in different forms.
  • Use equal gradients for parallel lines and negative reciprocal gradients for perpendicular lines.
  • Find equations of perpendicular bisectors and solve coordinate geometry problems.
Key Terms
Cartesian plane
A coordinate grid formed by perpendicular x- and y-axes.
Gradient
The vertical change divided by the horizontal change.
Midpoint
The point halfway between two endpoints.
Intercept
A point where a line crosses an axis.
Parallel
Having the same gradient and never meeting in the plane.
Perpendicular
Meeting at a right angle; non-vertical gradients multiply to -1.
Perpendicular bisector
A line that crosses a segment at its midpoint at 90 degrees.
What This Chapter Covers
  • Coordinates, gradient, length and midpoint
  • Equations in y=mx+c and general form
  • Lines through one or two points
  • Parallel and perpendicular lines
  • Perpendicular bisectors and mixed coordinate problems
Coordinates And Quadrants

A point is written as (x,y). Move horizontally using the x-coordinate and vertically using the y-coordinate. The signs of the coordinates determine the quadrant. Points on an axis do not lie in a quadrant.

Quadrant Sign of x Sign of y
I Positive Positive
II Negative Positive
III Negative Negative
IV Positive Negative
Gradient Of A Straight Line

The gradient measures rate of change. Use two points in the same order in both numerator and denominator. A positive gradient rises from left to right; a negative gradient falls.

m=\frac{y_2-y_1}{x_2-x_1}
Worked Example: Gradient

Question: Find the gradient of the line through A(-4,7) and B(6,-8).

  1. \Delta y=-8-7=-15.
  2. \Delta x=6-(-4)=10.
  3. m=-15/10.

Answer: m=-\frac32.

Length Of A Line Segment

The horizontal and vertical changes form the legs of a right-angled triangle, so the distance formula follows from Pythagoras’ theorem.

AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
Worked Example: Distance

Question: Find the distance between P(-2,3) and Q(4,11).

  1. Horizontal change =6.
  2. Vertical change =8.
  3. PQ=\sqrt{6^2+8^2}=\sqrt{100}.

Answer: PQ=10.

Midpoint

Average the x-coordinates and average the y-coordinates separately.

M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)
Worked Example: Midpoint

Question: Find the midpoint of (-5,9) and (7,-3).

  1. x-coordinate: (-5+7)/2=1.
  2. y-coordinate: (9-3)/2=3.

Answer: (1,3).

Equation Of A Straight Line

The form y=mx+c shows gradient and y-intercept directly. Other forms such as ax+by=c may need rearranging. A vertical line has equation x=k; its gradient is undefined. A horizontal line has equation y=k and gradient zero.

Worked Example: Read Gradient And Intercept

Question: Find the gradient and y-intercept of 5x+2y=14.

  1. Rearrange: 2y=-5x+14.
  2. Divide by 2: y=-\frac52x+7.

Answer: Gradient -5/2 and y-intercept 7.

Line Through A Point With A Known Gradient

Use y=mx+c and substitute the point to find c, or use point-gradient form.

y-y_1=m(x-x_1)
Worked Example: Point And Gradient

Question: Find the equation of the line with gradient 4 through (-2,5).

  1. Use y=4x+c.
  2. Substitute: 5=4(-2)+c.
  3. c=13.

Answer: y=4x+13.

Line Through Two Points

Find the gradient first, then use either point to find the intercept. Check that the other point satisfies the final equation.

Worked Example: Two Points

Question: Find the equation through (1,6) and (5,-2).

  1. m=(-2-6)/(5-1)=-8/4=-2.
  2. Use y=-2x+c.
  3. Substitute (1,6): 6=-2+c, so c=8.

Answer: y=-2x+8.

Parallel Lines

Parallel non-vertical lines have equal gradients. Their intercepts differ unless they are the same line.

Worked Example: Parallel Line

Question: Find the equation of the line parallel to 3x-2y=8 through (4,-1).

  1. Rearrange the given line: y=\frac32x-4, so m=3/2.
  2. Use y=\frac32x+c.
  3. Substitute (4,-1): -1=6+c, so c=-7.

Answer: y=\frac32x-7.

Perpendicular Lines

For non-vertical lines, perpendicular gradients are negative reciprocals. If one gradient is m, the other is -1/m, so their product is -1. Horizontal and vertical lines are perpendicular to each other.

m_1m_2=-1
Worked Example: Perpendicular Line

Question: Find the equation of the line perpendicular to y=\frac23x+5 through (6,1).

  1. The perpendicular gradient is -3/2.
  2. Use y=-\frac32x+c.
  3. Substitute (6,1): 1=-9+c, so c=10.

Answer: y=-\frac32x+10.

Perpendicular Bisectors

A perpendicular bisector passes through the midpoint of a segment and has a gradient perpendicular to that segment. Therefore, find the midpoint, find the original gradient, take its negative reciprocal, then form the equation.

Worked Example: Perpendicular Bisector

Question: Find the perpendicular bisector of the segment joining A(-3,8) and B(9,-2).

  1. Midpoint: (( -3+9)/2,(8-2)/2)=(3,3).
  2. Gradient of AB=(-2-8)/(9-(-3))=-10/12=-5/6.
  3. Perpendicular gradient =6/5.
  4. Use y-3=\frac65(x-3) and simplify.

Answer: y=\frac65x-\frac35, or 6x-5y=3.

Coordinate Proofs

Coordinates can be used to show that sides are equal, parallel or perpendicular. For example, equal distances can establish an isosceles triangle, equal gradients can establish parallel sides, and gradients with product -1 can establish a right angle.

Worked Example: Test Whether A Triangle Is Right-Angled

Question: Points are A(0,0), B(6,2) and C(2,8). Determine whether the triangle is right-angled and justify your conclusion.

  1. Gradient AB=2/6=1/3.
  2. Gradient AC=8/2=4.
  3. These gradients do not multiply to -1, so this proposed angle is not right; check the other vertices rather than forcing a conclusion.
  4. Gradient BA=1/3 and BC=(8-2)/(2-6)=6/(-4)=-3/2; product -1/2, also not right.
  5. The data do not form a right triangle, illustrating the need to calculate rather than assume from a sketch.

Answer: The triangle is not right-angled. A coordinate proof must follow the actual calculations.

Intercepts And Areas

The x-intercept is found by setting y=0; the y-intercept by setting x=0. These intercepts can form triangles with the coordinate axes, allowing area calculations.

Worked Example: Triangle With The Axes

Question: The line 2x+3y=12 meets the axes at A and B. Find the area of triangle OAB.

  1. At the x-axis, y=0, so x=6.
  2. At the y-axis, x=0, so y=4.
  3. The triangle has perpendicular base 6 and height 4.
  4. Area =\frac12(6)(4).

Answer: 12 square units.

Examination Guidance
  • Rearrange a line into y=mx+c before comparing gradients.
  • Keep the order of points consistent in both differences when calculating gradient.
  • Give line equations in a fully simplified form.
  • For perpendicular bisectors, use both the midpoint and the negative reciprocal gradient.
  • Check a line equation by substituting a known point.
Common Mistakes
  • Using (x_2-x_1)/(y_2-y_1) for gradient.
  • Averaging all four coordinates together instead of averaging x-values and y-values separately.
  • Assuming perpendicular gradients are merely negatives rather than negative reciprocals.
  • Trying to write a vertical line in y=mx+c form.
  • Using a diagram drawn not to scale as proof of parallelism or a right angle.
Knowledge Check And Practice

1. Find the gradient through (-1,4) and (3,12).

Answer: 2.

2. Find the midpoint of (8,-5) and (-2,7).

Answer: (3,1).

3. Find the distance between (1,2) and (7,10).

Answer: 10.

4. State the gradient of 4x+5y=20.

Answer: -4/5.

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

Answer: y=-3x+15.

6. Find the gradient of a line perpendicular to one with gradient 5/2.

Answer: -2/5.

7. Find the equation of the horizontal line through (4,-7).

Answer: y=-7.

8. Find the equation of the vertical line through (-3,5).

Answer: x=-3.
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