1.3 Coordinate geometry

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 y=mx+c for gradient/intercept, y−y₁=m(x−x₁) for a known point and gradient, and ax+by+c=0 for intersections, distances or a normal-vector description.

Convert between forms carefully and keep coefficients proportional when comparing the same line. For a line through two points, calculate its gradient before choosing the form.

2x−3y+6=0 becomes y=(2/3)x+2, revealing gradient 2/3 and y-intercept 2.

Multiplying all coefficients by a non-zero constant gives the same line; changing only one coefficient changes its geometry.

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.

Substitute the line equation into the circle equation. The resulting quadratic has two roots for a secant, one repeated root for a tangent, and no real roots when the line misses the circle.

The repeated-root condition is equivalent to the perpendicular distance from the centre to the line equalling the radius. Use whichever route makes the parameter condition clearer.

Substituting y=mx+c into a circle gives Δ=0 at tangency; the repeated x-value then gives the contact point.

A discriminant classifies real algebraic intersections, but a point must also satisfy any stated segment or domain restriction.

An intersection of y=f(x) and y=g(x) has f(x)=g(x); solve for x, then substitute to find y. The number of real roots gives the number of intersections.

Factorisation may reveal exact roots; otherwise use a graph or numerical method only after locating intervals. Reject roots outside the domain or interval shown.

For y=x² and y=2x+3, x²=2x+3 gives x=3 or −1, producing points (3,9) and (−1,1).

Solving f(x)=g(x) finds x-coordinates only; reporting roots without their corresponding y-values is not a complete intersection answer.