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

Coordinate Geometry (Pure Mathematics 1 – Cambridge 9709)

1. Functions – a quick refresher

2. Quadratics

2.1 Standard, factorised and completed‑square forms

• Standard form $ax^{2}+bx+c=0\quad(aeq0)$
• Factorised form $a(x-r_{1})(x-r_{2})=0$
• Completed‑square form $a\bigl(x-h\bigr)^{2}+k=0$ where $h=-\dfrac{b}{2a},\;k=\dfrac{b^{2}-4ac}{4a}$

2.2 Solving a quadratic

1. Factorise (if possible).
2. Complete the square: $$ax^{2}+bx+c=0\;\Longrightarrow\;(x+\tfrac{b}{2a})^{2}=\tfrac{b^{2}-4ac}{4a^{2}}$$
3. Quadratic formula: $$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$

2.3 Discriminant and nature of roots

The discriminant $\Delta=b^{2}-4ac$ tells us how many real roots the quadratic has:

Discriminant $\Delta$	Number of real roots	Root type
$\Delta>0$	2 distinct real roots	$x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$
$\Delta=0$	1 real double root	$x=-\dfrac{b}{2a}$
$\Delta<0$	0 real roots (2 complex conjugates)	Complex roots

2.4 Quadratic inequalities

To solve $ax^{2}+bx+c\;{\color{blue}{\lessgtr}}\;0$:

1. Find the real roots $r_{1},r_{2}$ (if any) using the discriminant.
2. Sketch a simple sign chart (parabola opens upwards if $a>0$, downwards if $a<0$).
3. Read off the intervals where the expression satisfies the required inequality.

Example. Solve $x^{2}-4x-5\ge0$.

1. Roots: $x=\dfrac{4\pm\sqrt{16+20}}{2}=5,\;-1$.
2. Since $a=1>0$, the parabola is upward‑opening; it is ≥0 outside the roots.
3. Solution: $x\le-1\;\text{or}\;x\ge5$.

2.5 Quadratic in a function of $x$

Sometimes the variable appears as a function, e.g. $x^{4}-5x^{2}+4=0$. Set $u=x^{2}$, solve the resulting quadratic in $u$, then back‑substitute.

Example. $x^{4}-5x^{2}+4=0$

1. Let $u=x^{2}$ → $u^{2}-5u+4=0$.
2. Factor: $(u-1)(u-4)=0$ → $u=1$ or $u=4$.
3. Back‑substitute: $x^{2}=1$ or $x^{2}=4$ → $x=\pm1,\;\pm2$.

2.6 Simultaneous linear + quadratic equations

Typical A‑Level question: solve the system

\[ \begin{cases} y=mx+c\\[2mm] ax^{2}+bx+c=0 \end{cases} \]

Substitute the linear expression for $y$ into the quadratic, solve the resulting quadratic in $x$, then obtain $y$.

Worked example. Solve

\[ \begin{cases} y=2x-3\\[2mm] x^{2}+y^{2}=25 \end{cases} \]

1. Substitute $y$: $x^{2}+(2x-3)^{2}=25$.
2. Expand: $x^{2}+4x^{2}-12x+9=25\;\Longrightarrow\;5x^{2}-12x-16=0$.
3. Quadratic formula: $x=\dfrac{12\pm\sqrt{144+320}}{10} =\dfrac{12\pm\sqrt{464}}{10} =\dfrac{12\pm4\sqrt{29}}{10} =\dfrac{6\pm2\sqrt{29}}{5}$.
4. Corresponding $y$ from $y=2x-3$ gives the two intersection points.

3. Straight lines

3.1 Gradient (slope) and the point‑slope form

For a non‑vertical line the gradient $m$ is

$$m=\frac{\Delta y}{\Delta x}= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

With a known point $(x_{1},y_{1})$:

$$y-y_{1}=m\,(x-x_{1})\tag{1}$$

3.2 Slope‑intercept form

If the $y$‑intercept $c$ is known (point where $x=0$):

$$y=mx+c\tag{2}$$

3.3 General (standard) form

Multiplying (1) or (2) to clear fractions gives

$$Ax+By+C=0\tag{3}$$

where $A$, $B$, $C$ are not all zero. Gradient $m=-\dfrac{A}{B}$ (provided $Beq0$). For a vertical line $B=0$ and the equation reduces to $x=k$.

3.4 Parallel and perpendicular lines

• Parallel: $m_{1}=m_{2}$ or $A_{1}B_{2}=A_{2}B_{1}$.
• Perpendicular: $m_{1}m_{2}=-1$ or $A_{1}A_{2}+B_{1}B_{2}=0$.

3.5 Intersection of two lines

Given

\[ \begin{cases} A_{1}x+B_{1}y+C_{1}=0\\[2mm] A_{2}x+B_{2}y+C_{2}=0 \end{cases} \]

the determinant $\Delta=A_{1}B_{2}-A_{2}B_{1}$ determines the relationship:

• Unique intersection ($\Deltaeq0$): $$x=\frac{B_{1}C_{2}-B_{2}C_{1}}{\Delta},\qquad y=\frac{C_{1}A_{2}-C_{2}A_{1}}{\Delta}$$
• Parallel ($\Delta=0$ and $A_{1}C_{2}eq A_{2}C_{1}$).
• Coincident ($\Delta=0$ and $A_{1}C_{2}=A_{2}C_{1}$, equivalently the two equations are proportional).

3.6 Solving a linear‑quadratic system (link to §2.6)

Use substitution (replace $y$ from the linear equation in the quadratic) and then apply the methods of §2.

4. Mid‑point and distance

4.1 Mid‑point formula

$$M\Bigl(\frac{x_{1}+x_{2}}{2},\;\frac{y_{1}+y_{2}}{2}\Bigr)$$

4.2 Distance formula

$$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$$

5. Circles

5.1 Standard (centre‑radius) form

$$ (x-h)^{2}+(y-k)^{2}=r^{2}\tag{4}$$

Centre $C(h,k)$, radius $r>0$.

5.2 General (expanded) form

$$x^{2}+y^{2}+Dx+Ey+F=0\tag{5}$$

Relation to centre and radius:

$$h=-\frac{D}{2},\qquad k=-\frac{E}{2},\qquad r=\sqrt{h^{2}+k^{2}-F}$$

5.3 Completing the square (from the general form)

\[ \begin{aligned} x^{2}+Dx+y^{2}+Ey &= -F\\[2mm] \left(x+\frac{D}{2}\right)^{2}+\left(y+\frac{E}{2}\right)^{2} &= \frac{D^{2}+E^{2}}{4}-F \end{aligned} \]

The right‑hand side must be positive for a real circle.

5.4 Intersection of a line and a circle

1. Write the line in either $y=mx+c$ or $x=ky+d$ form.
2. Substitute the linear expression for the missing variable into (4) or (5).
3. Simplify → a quadratic in the remaining variable.
4. Solve the quadratic (use discriminant to decide 0, 1 or 2 points).
5. Back‑substitute to obtain the corresponding coordinates.

Example – line meeting a circle

Find the points where the line $y=2x+1$ cuts the circle $(x-3)^{2}+(y+2)^{2}=25$.

1. Substitute $y$: $$(x-3)^{2}+\bigl(2x+1+2\bigr)^{2}=25$$
2. Simplify: $$(x-3)^{2}+(2x+3)^{2}=25$$ $$x^{2}-6x+9+4x^{2}+12x+9=25$$ $$5x^{2}+6x-7=0$$
3. Quadratic formula: $$x=\frac{-6\pm\sqrt{6^{2}-4\cdot5(-7)}}{2\cdot5} =\frac{-6\pm\sqrt{176}}{10} =\frac{-3\pm2\sqrt{11}}{5}$$
4. Corresponding $y$ from $y=2x+1$ gives the two intersection points: $$\Bigl(\tfrac{-3+2\sqrt{11}}{5},\; \tfrac{-1+4\sqrt{11}}{5}\Bigr),\qquad \Bigl(\tfrac{-3-2\sqrt{11}}{5},\; \tfrac{-1-4\sqrt{11}}{5}\Bigr).$$

5.5 Tangents from an external point (optional extension)

If $P(p,q)$ lies outside the circle $(x-h)^{2}+(y-k)^{2}=r^{2}$, the length of the tangent $PT$ is

$$PT=\sqrt{(p-h)^{2}+(q-k)^{2}-r^{2}}$$

Equating this length to the distance from $P$ to a point on the line yields the required tangent equations.

6. Circular measure

7. Trigonometry basics (relevant to coordinate geometry)

8. Summary of key formulae

Concept	Formula	Comments
Gradient (slope)	$m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$	Undefined for vertical lines.
Point‑slope equation	$y-y_{1}=m(x-x_{1})$	Any known point may be used.
Slope‑intercept equation	$y=mx+c$	$c$ is the $y$‑intercept.
General linear equation	$Ax+By+C=0$	Gradient $m=-A/B$ (if $Beq0$).
Parallel lines	$m_{1}=m_{2}$	Or $A_{1}B_{2}=A_{2}B_{1}$.
Perpendicular lines	$m_{1}m_{2}=-1$	Or $A_{1}A_{2}+B_{1}B_{2}=0$.
Intersection of two lines	$x=\dfrac{B_{1}C_{2}-B_{2}C_{1}}{A_{1}B_{2}-A_{2}B_{1}},\; y=\dfrac{C_{1}A_{2}-C_{2}A_{1}}{A_{1}B_{2}-A_{2}B_{1}}$	Denominator $eq0$ for a unique point.
Mid‑point	$\bigl(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2}\bigr)$
Distance	$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
Quadratic discriminant	$\Delta=b^{2}-4ac$	Determines number & type of real roots.
Circle (standard)	$(x-h)^{2}+(y-k)^{2}=r^{2}$	Centre $(h,k)$, radius $r$.
Circle (general)	$x^{2}+y^{2}+Dx+Ey+F=0$	Centre $(-D/2,-E/2)$, $r=\sqrt{(D^{2}+E^{2})/4-F}$.
Radian–degree conversion	$1\;\text{rad}=180^{\circ}/\pi,\;\;1^{\circ}=\pi/180\;\text{rad}$
Arc length	$\ell=r\theta$ (θ in radians)
Sector area	$A_{\text{sector}}=\frac12 r^{2}\theta$