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Mathematics · 13 / 23

Coordinate geometry: midpoint, distance and line

High school foundations · Worked example and 4 exercises with solutions, at your pace.

Need a foundation?Pythagoras: lengths and right-angle checksVectors: displacement, not location
UnderstandSee an exampleGuided exercise3 progressive challenges

The principle, simply

In an orthonormal frame, distance is the norm of coordinate differences, by Pythagoras. A midpoint averages corresponding coordinates. A nonvertical line is y=mx+p with m=Δy/Δx. A vertical line x=constant has no finite real slope.

d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
THE PICTURE TO REMEMBERCoordinate geometry: midpoint, distance and line
  1. 01Coordinates
  2. 02Differences or averages
  3. 03Matching formula
  4. 04Frame conditions

WORKED EXAMPLE

From principle to reasoning

A(1,2), B(5,5) give vector (4,3), distance 5 and midpoint (3,3.5). Slope is 3/4 and intercept 5/4.

Solve a problem in this subject

  1. Define the objects

    State unknowns, domain and data. Real square roots need nonnegative arguments; denominators must be nonzero.

  2. Choose a transformation

    Name the rule: expansion, equivalent equations, factoring, differentiation. Check its conditions. Squaring an equation requires testing candidates in the original.

  3. Show the reasoning

    Each line should follow from the previous one. Keep exact values where possible and distinguish equality, approximation and implication.

  4. Test and conclude

    Substitute into the original expression. Check graph points and domain boundaries. State a solution set or the required unit.

YOUR TURN

01 · Try before checking

Find the midpoint of A(−2,4) and B(6,0).

One hint

Average each coordinate separately.

Reveal and understand the solution

((−2+6)/2,(4+0)/2)=(2,2), halfway in both coordinates.

Compare method, conditions and answer; identify the exact part to revisit.

IMPROVE THROUGH PRACTICE

Three additional challenges with solutions

Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.

Reflect after practising
02 · Recognise and understandLink to this exercise

Find the midpoint of A(−2,1) and B(4,5).

A hint to get started

Average each pair of coordinates.

Step-by-step solution
  1. xM = (−2 + 4)/2 = 1.
  2. yM = (1 + 5)/2 = 3.
  3. M(1,3); this is not vector AB.

Which step did you miss? Retry without help, then explain why the method works.

03 · Apply the methodLink to this exercise

Find the line through A(1,3) and B(3,7).

A hint to get started

Calculate its slope, then use one point.

Step-by-step solution
  1. Slope: (7 − 3)/(3 − 1) = 2.
  2. 3 = 2 × 1 + b, so b = 1.
  3. The line is y = 2x + 1; B also satisfies it.

Which step did you miss? Retry without help, then explain why the method works.

04 · Go furtherLink to this exercise

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

A hint to get started

At the intersection the y expressions are equal.

Step-by-step solution
  1. 2x + 1 = −x + 7.
  2. 3x = 6 gives x = 2 and y = 5.
  3. The intersection is (2,5), satisfying both lines.

Which step did you miss? Retry without help, then explain why the method works.

The trap to avoid

The distance formula needs an orthonormal frame; averaging distances from the origin is not a midpoint.

Without notes, can you…

  • Explain the principle in your own words?
  • Rebuild the diagram and its connections?
  • Solve the exercise and justify the method?

Choosing a state saves it in this browser. This is your assessment, not a grade.

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