Intersections, collinear points and perpendicular bisectors

Where two lines cross

• Two non-parallel lines meet at a single point of intersection — the $(x, y)$ that satisfies both equations at once.
• Solve the two equations simultaneously: set them equal (if both are $y = \ldots$), solve for $x$, then substitute back for $y$.
• The point of intersection lies on both lines, so it works in either equation — use the other one to check.

Testing whether points are collinear

• Collinear points all lie on one straight line.
• Points are collinear when the gradient between each pair is the same. Find $m$ for two of the pairs (sharing a common point) and compare.
• Equal gradients through a shared point ⇒ collinear; different gradients ⇒ they form a triangle.

The perpendicular bisector

• The perpendicular bisector of a segment $AB$ is the line that cuts $AB$ in half and crosses it at a right angle.
• Every point on it is the same distance from $A$ as from $B$ — useful for "equidistant" problems (e.g. the fairest place to build between two towns).
• Build it in two steps:
  • Midpoint of $AB$ — the point it must pass through.
  • Negative reciprocal of the gradient of $AB$ — its gradient.
• Then use point–gradient form with that midpoint and gradient.