1.3 Coordinate geometry
- Syllabus
- 9709–2028–2029
- Topic
- 1.3
- Level
- AS
Learning objectives
1.3.1Straight lines• find the equation of a straight line given sufficient information e.g. given two points, or one point and the gradient.1.3.2Line equation forms• interpret and use any of the forms y = mx + c, y - y1 = m(x - x1), ax + by + c = 0 in solving problems Including calculations of distances, gradients, midpoints, points of intersection and use of the relationship between the gradients of parallel and perpendicular lines.1.3.3Circle equation• understand that the equation (x - a)2 + (y - b)2 = r 2 represents the circle with centre (a, b) and radius r Including use of the expanded form x 2 + y 2 + 2gx + 2fy + c = 0.1.3.4Lines and circles• use algebraic methods to solve problems involving lines and circles Including use of elementary geometrical properties of circles, e.g. tangent perpendicular to radius, angle in a semicircle, symmetry. Implicit differentiation is not included.1.3.5Graph intersections• understand the relationship between a graph and its associated algebraic equation, and use the relationship between points of intersection of graphs and solutions of equations. e.g. to determine the set of values of k for which the line y = x + k intersects, touches or does not meet a quadratic curve.
A straight line is determined by gradient and one point
The gradient m measures change in y per unit change in x. A line can be written y=mx+c, y−y₁=m(x−x₁), or ax+by+c=0 depending on the information given.
Calculate m=(y₂−y₁)/(x₂−x₁), then substitute a known point to determine the intercept. Parallel lines have equal gradients; perpendicular non-vertical lines satisfy m₁m₂=−1.
Through (2,5) with gradient −3: y−5=−3(x−2), so y=−3x+11.
The intercept is not the y-coordinate of every point, and a vertical line has undefined gradient rather than gradient zero.
Use one coordinate toolkit to build lines, bisectors and intersections
For $A(x_1,y_1)$ and $B(x_2,y_2)$:m_{AB}=\frac{y_2-y_1}{x_2-x_1},\quad M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right),\quad AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
Use y=mx+c to expose gradient/intercept, y−y1=m(x−x1) with a known point, and ax+by+c=0 when an integer-coefficient final form or simultaneous intersection is convenient.
Parallel non-vertical lines: $m_1=m_2$. Perpendicular non-vertical lines: $m_1m_2=-1$. Find an intersection by solving both line equations simultaneously.
A perpendicular bisector passes through midpoint M and has gradient −1/mAB. Compute those two facts first, then use point-gradient form; this is the standard route to a circle centre from a chord.
The negative reciprocal rule needs the minus sign and reciprocal. Handle horizontal/vertical pairs separately, and remember that multiplying every coefficient of ax+by+c=0 by the same non-zero constant gives the same line.
A circle equation encodes its centre and radius
The circle with centre (a,b) and radius r has (x−a)²+(y−b)²=r². Expanding gives a general quadratic form, but completing squares recovers the geometry.
Use the radius as a distance, check r²>0, and substitute a point to test whether it lies on the circle. A zero radius is a single point, not a proper circle.
x²+y²−6x+4y−3=0 becomes (x−3)²+(y+2)²=16, so centre (3,−2) and radius 4.
The signs in the centre reverse when completing squares: (x−3)² gives centre x=3, while (y+2)² gives y=−2.
Choose algebra or circle geometry for line-circle problems
Substitute the line equation into the circle equation. The resulting quadratic has two real roots for a secant, one repeated root for a tangent and no real roots when the line misses the circle.
Use elementary properties when they shorten the route: a radius is perpendicular to the tangent at contact; the perpendicular bisector of a chord passes through the centre; an angle in a semicircle is 90∘; symmetry about a diameter pairs points.
For a tangent through a known point, either impose discriminant =0 after substitution or find the radius gradient and use its negative reciprocal. Then substitute back to obtain and verify the contact point.
A line is tangent exactly when the perpendicular distance from the circle centre to the line equals the radius. This may classify a parameter without solving both coordinates.
A repeated root proves tangency to the infinite line, but the point must still satisfy any stated segment/domain condition. Keep radius, diameter and chord roles distinct.
A parameter controls how many times two graphs meet
Intersections of y=f(x) and y=g(x) satisfy f(x)=g(x). Real roots are intersection x-coordinates; substitute each back to obtain the matching y-coordinate.
If one graph contains a parameter k, rearrange the intersection equation into a polynomial in x whose coefficients depend on k. For a quadratic, use its discriminant to classify the graph positions.
For $Ax^2+B(k)x+C(k)=0$: $\Delta(k)>0$: two intersections; $\Delta(k)=0$: one repeated intersection/touch; $\Delta(k)<0$: no real intersection.
$y=x^2$ and $y=x+k$ meet where $x^2-x-k=0$. Here $\Delta=1+4k$, so they meet twice if $k>-\tfrac14$, touch if $k=-\tfrac14$, and do not meet if $k<-\tfrac14$.
Classify the whole requested set of k values and preserve strict/equality signs. Root count describes intersections only within any stated domain or interval.