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P1.2 - Coordinate geometry in the (x, y) plane

Syllabus
2019
Topic
P1.2
Level
AS

Learning objectives

P1.2.1Equation of a straight lineEquation of a straight line, To include: including the forms (i) the equation of a line through two given points y − y = m(x − x) and 1 1 (ii) the equation of a line parallel (or perpendicular) to a ax + by + c = 0. given line through a given point.; For example, the line perpendicular to the line 3x + 4y = 18 through the point (2, 3) has equation y − 3 = 4 (x − 2).P1.2.2Parallel and perpendicular linesConditions for two straight lines to be parallel or perpendicular to each other.

Build a straight-line equation from a point and gradient

m=y2−y1x2−x1,y−y1=m(x−x1),ax+by+c=0m=\frac{y_2-y_1}{x_2-x_1},\qquad y-y_1=m(x-x_1),\qquad ax+by+c=0

A straight line has constant gradient mm. Two points determine that gradient when x2≠x1x_2\ne x_1; substituting either point into y−y1=m(x−x1)y-y_1=m(x-x_1) then fixes the unique line. Expand and collect terms only if a requested form such as ax+by+c=0ax+by+c=0 is needed.

Required line Gradient to use
through two points m=(y2−y1)/(x2−x1)m=(y_2-y_1)/(x_2-x_1)
parallel through a point the given line's gradient
perpendicular through a point the negative reciprocal of the given gradient

For the line perpendicular to 3x+4y=183x+4y=18 through (2,3)(2,3), first write the given line as y=−34x+92y=-\tfrac34x+\tfrac92. Its perpendicular gradient is 43\tfrac43, so the required equation is y−3=43(x−2)y-3=\tfrac43(x-2). Equivalently, 4x−3y+1=04x-3y+1=0. Substituting (2,3)(2,3) checks the point lies on it.

The gradient formula is undefined when x2=x1x_2=x_1; that line is vertical and has equation x=x1x=x_1. Do not take a negative reciprocal when constructing a parallel line, and do not lose the given point while changing equation form.

Use gradients to test parallel and perpendicular lines

Gradient records a line's direction. Lines with the same direction are parallel; a quarter-turn changes a non-zero finite gradient to its negative reciprocal.

Relationship Condition for non-vertical lines
parallel m1=m2m_1=m_2
perpendicular m1m2=−1m_1m_2=-1, equivalently m2=−1/m1m_2=-1/m_1

A line through P(−3,7)P(-3,7) and Q(9,11)Q(9,11) has gradient 11−79−(−3)=13\tfrac{11-7}{9-(-3)}=\tfrac13. A line through Q(9,11)Q(9,11) and R(12,2)R(12,2) has gradient 2−1112−9=−3\tfrac{2-11}{12-9}=-3. Their product is −1-1, so the lines are perpendicular and ∠PQR=90∘\angle PQR=90^\circ.

For parallel lines, equal gradients establish equal direction. To show they are distinct rather than the same line, compare their intercepts or test whether a point on one lies on the other.

The product rule applies only when both gradients are defined. Vertical lines x=ax=a are parallel to other vertical lines and perpendicular to horizontal lines y=by=b. Opposite signs alone do not prove perpendicularity; the magnitudes must also be reciprocal.