Intersecting lines and applications
Finding where two lines meet
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The point of intersection is the point that lies on both lines, so its coordinates satisfy both equations.
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Finding it means solving the two equations simultaneously.
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The two standard methods:
- Substitution — best when at least one equation is already
- Elimination — best when both are in general form
Solving by substitution
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Make the subject of one equation (or use it if already done).
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Substitute that expression into the other equation.
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Solve for .
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Substitute back to find .
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State the answer as a point, .
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Worked through — where do and meet?
- Set them equal:
- , so
- Substituting:
- Intersection:
Solving by elimination
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Multiply one or both equations so that the coefficients of one variable match.
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Add or subtract to eliminate that variable.
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Solve, then substitute back.
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Worked through — where do and meet?
- The coefficients are already and , so add the equations: , giving
- Substituting into the second: , so
- Intersection:
Always check
- Substitute the point into BOTH original equations. A point that satisfies only one is not an intersection, and this check catches every arithmetic slip.
When there is no intersection
- Parallel lines never meet, so there is no point of intersection.
- Algebraically, solving gives something impossible such as .
- Say what it means: "The lines are parallel and therefore do not intersect."
When there are infinitely many
- If the two equations describe the same line, every point on it is an intersection.
- Algebraically, solving gives something always true such as .
- Check by rearranging both to — if they match exactly, they are the same line.
Applying the whole toolkit
- Most Merit and Excellence tasks combine several methods. The common jobs:
| Question asks | Method |
|---|---|
| Is the triangle isosceles? | Distance — compare side lengths |
| Is there a right angle? | Gradients — check |
| Is it a parallelogram? | Midpoints of the diagonals, or gradients of opposite sides |
| Is it a rhombus? | Parallelogram and all sides equal |
| Is it a rectangle? | Parallelogram and a right angle |
| Where do two paths cross? | Simultaneous equations |
| What is the perimeter? | Distance for each side, then add |
- Read the question to decide which tool it needs. That decision is much of the assessed skill.
Relating findings to the context
- At Merit, the standard requires "relating findings to a context".
- So do not stop at "". Write: "the two paths meet at a right angle, so the corner of the reserve is square".
- At Excellence, look for a generalisation — a result that holds for a whole family of cases, not just the numbers given.
Worked ExampleCombining methods in a context
A park is bounded by four straight paths with corners at , , and , in metres. Determine whether the corner at is a right angle, find the length of path , and find where the diagonals and cross.
Step 1 — Decide which tools are needed
- Right angle at → compare the gradients of the two paths meeting at
- Length of → the distance formula
- Where the diagonals cross → the midpoint of each, or simultaneous equations
Step 2 — Test the corner at
The two paths meeting at are and .
Gradient of :
Gradient of :
Multiply them:
Step 3 — State the conclusion in context
Since the product of the gradients is , the two paths are perpendicular.
Step 4 — Find the length of
Simplify the surd:
Step 5 — Find where the diagonals cross
The diagonals of join opposite corners: and .
Find the equation of , from to :
It passes through the origin, so:
Find the equation of , from to :
Both endpoints have , so this diagonal is the horizontal line:
Step 6 — Solve simultaneously
Substitute into the first equation:
Step 7 — Check the point on both lines
- On : ✓
- On : ✓
Step 8 — A further observation
Is the midpoint of either diagonal?
Both midpoints are — the same as the intersection point. So the diagonals bisect each other, which means is a parallelogram.
Combined with the right angle at found in Step 3, is a rectangle — the paths enclose a rectangular park.