Coordinate geometry: equations of straight lines, intersections, midpoints, gradients, circles

Equations of Straight Lines

Think of a straight line as a road that never turns. Its equation tells you exactly where the road lies on the map.

• Gradient (slope) – how steep the road is: $m = \dfrac{y_2-y_1}{x_2-x_1}$
• Point–Slope form – starts from a known point on the road: $y-y_1 = m(x-x_1)$
• Slope–Intercept form – shows where the road crosses the y‑axis: $y = mx + c$

Example: Find the equation of the line through $(2,3)$ and $(5,11)$.

1. Compute the gradient: $m = \dfrac{11-3}{5-2} = \dfrac{8}{3}$
2. Use point–slope with point $(2,3)$: $y-3 = \dfrac{8}{3}(x-2)$
3. Simplify to slope–intercept: $y = \dfrac{8}{3}x - \dfrac{1}{3}$

So the road is described by $y = \dfrac{8}{3}x - \dfrac{1}{3}$ 📏.

Intersections of Lines

When two roads cross, the intersection point is where their equations agree.

Example: Find the intersection of $y = 2x + 1$ and $y = -x + 4$.

1. Set the right‑hand sides equal: $2x + 1 = -x + 4$
2. Combine like terms: $3x = 3$
3. Find $x$: $x = 1$
4. Substitute back: $y = 2(1) + 1 = 3$

The roads meet at $(1,\,3)$ 🎯.

Midpoints & Gradients

Midpoints are the “centre of the road” between two points, and gradients help you find perpendicular roads.

Midpoint formula:

$M = \left(\dfrac{x_1+x_2}{2},\,\dfrac{y_1+y_2}{2}\right)$

Example: Midpoint of $(1,2)$ and $(5,6)$:

1. $x$‑coordinate: $\dfrac{1+5}{2} = 3$
2. $y$‑coordinate: $\dfrac{2+6}{2} = 4$

Midpoint is $(3,\,4)$ 🧮.

Perpendicular gradients – If one road has slope $m$, a road perpendicular to it has slope $-\dfrac{1}{m}$.

Example: Road with slope $3$ → perpendicular slope $-\dfrac{1}{3}$.

Circles

A circle is all the points that are the same distance from a centre.

Standard form:

$(x-h)^2 + (y-k)^2 = r^2$

Example: Centre at $(3,\,-2)$, radius $5$:

$(x-3)^2 + (y+2)^2 = 25$

Expanding gives the general form:

$x^2 + y^2 - 6x + 4y - 12 = 0$

Intersection with a line – substitute the line’s equation into the circle’s.

Example: Line $y = x$ and circle centred at origin with radius $5$:

$x^2 + x^2 = 25 \;\Rightarrow\; 2x^2 = 25 \;\Rightarrow\; x = \pm\sqrt{\dfrac{25}{2}}$

Points of intersection: $\left(\pm\sqrt{\dfrac{25}{2}},\,\pm\sqrt{\dfrac{25}{2}}\right)$ 🧠.

Exam Tips & Tricks 🎯

• Always check your algebra – a small sign error can change the answer.
• When solving for intersections, substitute rather than solve simultaneously if one equation is simpler.
• For circles, complete the square to move from general to standard form.
• Remember the negative reciprocal rule for perpendicular lines.
• When dealing with midpoints, verify by plugging back into the original points.
• Use the gradient formula to check if two lines are parallel (same slope) or perpendicular (product of slopes = -1).
• Practice graphing a few examples to visualise the relationships.

Good luck – you’ve got this! 🚀