Coordinate Geometry

# Coordinate Geometry ## Objectifs et prérequis This chapter aims to develop students’ ability to apply algebraic methods to geometric problems using the Cartesian coordinate system. Students will learn to calculate distances, midpoints, and division points; determine gradients and equations of lines; analyse intersections, angles, and parallelism/perpendicularity; compute point-to-line distances; reduce non-linear relationships to linear form for graphing; and interpret loci in Cartesian and parametric forms. Prerequisites: Proficiency in algebraic manipulation, solving linear equations, basic trigonometry (for angle between lines), and familiarity with the Cartesian plane (plotting points, quadrants). ## Introduction Coordinate geometry, or analytic geometry, bridges algebra and geometry by representing geometric figures through coordinates and equations. This enables precise calculation of properties such as length, slope, and intersection without relying solely on diagrammatic reasoning. In the Cameroonian Anglophone Lower Sixth curriculum, this chapter builds on foundational knowledge of graphs and functions to equip students with tools essential for calculus, physics, and engineering applications. Mastery requires not only formula recall but also geometric interpretation and problem-solving flexibility. ## I. Distance between two points and Midpoint of a line segment ### Definition and formula The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the Cartesian plane is derived from the Pythagorean theorem. > The distance $d$ between $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by: > $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ The midpoint $M$ of segment $AB$ is the point equidistant from $A$ and $B$. > The coordinates of the midpoint $M$ of $A(x_1, y_1)$ and $B(x_2, y_2)$ are: > $$M\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$ ### Properties - Distance is always non-negative; $d = 0$ iff $A \equiv B$. - The midpoint formula is a special case of the section formula with ratio $1:1$. - Distance is invariant under translation. ### Solved example 1 Find the distance between $P(3, -2)$ and $Q(-1, 4)$. $$ \begin{align*} d &= \sqrt{(-1 - 3)^2 + (4 - (-2))^2} \\ &= \sqrt{(-4)^2 + (6)^2} \\ &= \sqrt{16 + 36} \\ &= \sqrt{52} \\ &= 2\sqrt{13} \end{align*} $$ Answer: $2\sqrt{13}$ units. ### Solved example 2 Find the midpoint of $R(-4, 5)$ and $S(6, -3)$. $$ \begin{align*} M_x &= \frac{-4 + 6}{2} = \frac{2}{2} = 1 \\ M_y &= \frac{5 + (-3)}{2} = \frac{2}{2} = 1 \end{align*} $$ Answer: $M(1, 1)$. ### Application exercises (corrected) 1. Calculate the distance between $A(0, 0)$ and $B(5, 12)$. Correction: $d = \sqrt{(5-0)^2 + (12-0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$. 2. Find the midpoint of $C(-7, 2)$ and $D(3, -8)$. Correction: $M_x = \frac{-7+3}{2} = -2$, $M_y = \frac{2+(-8)}{2} = -3$ → $M(-2, -3)$. 3. If the midpoint of $E(2, y)$ and $F(8, 6)$ is $G(5, 4)$, find $y$. Correction: $\frac{y + 6}{2} = 4 \Rightarrow y + 6 = 8 \Rightarrow y = 2$. ## II. Coordinates of a point that divides a line internally and externally ### Definition and formula A point $P$ divides segment $AB$ in the ratio $m:n$. - **Internal division**: $P$ lies between $A$ and $B$. - **External division**: $P$ lies on the line $AB$ extended beyond $A$ or $B$. > For internal division in ratio $m:n$, coordinates of $P$ dividing $A(x_1, y_1)$ and $B(x_2, y_2)$: > $$P\left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)$$ > For external division in ratio $m:n$ ($m \neq n$), coordinates of $P$: > $$P\left( \frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n} \right)$$ ### Properties - If $m = n$, internal division gives the midpoint. - External division is undefined if $m = n$ (denominator zero). - The ratio $m:n$ can be written as $\lambda:1$ where $\lambda = m/n$. ### Solved example 1 (internal) Find coordinates of point dividing $A(1, 2)$ and $B(7, 8)$ internally in ratio $2:3$. $$ \begin{align*} x &= \frac{2 \cdot 7 + 3 \cdot 1}{2+3} = \frac{14 + 3}{5} = \frac{17}{5} = 3.4 \\ y &= \frac{2 \cdot 8 + 3 \cdot 2}{5} = \frac{16 + 6}{5} = \frac{22}{5} = 4.4 \end{align*} $$ Answer: $P\left( \frac{17}{5}, \frac{22}{5} \right)$. ### Solved example 2 (external) Find coordinates of point dividing $A(-2, 4)$ and $B(4, -2)$ externally in ratio $3:1$. $$ \begin{align*} x &= \frac{3 \cdot 4 - 1 \cdot (-2)}{3-1} = \frac{12 + 2}{2} = \frac{14}{2} = 7 \\ y &= \frac{3 \cdot (-2) - 1 \cdot 4}{2} = \frac{-6 - 4}{2} = \frac{-10}{2} = -5 \end{align*} $$ Answer: $P(7, -5)$. ### Application exercises (corrected) 1. Find point dividing $J(0, 0)$ and $K(9, 3)$ internally in ratio $1:2$. Correction: $x = \frac{1\cdot9 + 2\cdot0}{3} = 3$, $y = \frac{1\cdot3 + 2\cdot0}{3} = 1$ → $P(3, 1)$. 2. Find point dividing $L(-3, -1)$ and $M(5, 7)$ externally in ratio $2:1$. Correction: $x = \frac{2\cdot5 - 1\cdot(-3)}{2-1} = \frac{10+3}{1} = 13$, $y = \frac{2\cdot7 - 1\cdot(-1)}{1} = 14+1=15$ → $P(13, 15)$. 3. If $P(4, 5)$ divides $A(2, 1)$ and $B(x, 9)$ internally in ratio $1:2$, find $x$. Correction: $4 = \frac{1\cdot x + 2\cdot 2}{3} \Rightarrow 12 = x + 4 \Rightarrow x = 8$. ## III. Gradient of a line and collinear points ### Definition and formula The gradient (slope) $m$ of a line measures its steepness and direction. > The gradient of line through $A(x_1, y_1)$ and $B(x_2, y_2)$ is: > $$m = \frac{y_2 - y_1}{x_2 - x_1}, \quad (x_2 \neq x_1)$$ ### Properties - $m > 0$: line rises left to right; $m < 0$: falls; $m = 0$: horizontal; undefined: vertical. - Parallel lines have equal gradients. - Three or more points are collinear iff the slope between any two pairs is equal. ### Solved example 1 Find gradient of line through $P(-1, 3)$ and $Q(4, -2)$. $$ m = \frac{-2 - 3}{4 - (-1)} = \frac{-5}{5} = -1 $$ Answer: $m = -1$. ### Solved example 2 Show that $A(1, 2)$, $B(3, 6)$, $C(5, 10)$ are collinear. $$ m_{AB} = \frac{6-2}{3-1} = \frac{4}{2} = 2, \quad m_{BC} = \frac{10-6}{5-3} = \frac{4}{2} = 2 $$ Since $m_{AB} = m_{BC}$, points are collinear. ### Application exercises (corrected) 1. Find gradient of line through $X(0, -4)$ and $Y(6, 2)$. Correction: $m = \frac{2 - (-4)}{6 - 0} = \frac{6}{6} = 1$. 2. Determine if $D(-2, 1)$, $E(0, 3)$, $F(2, 5)$ are collinear. Correction: $m_{DE} = \frac{3-1}{0-(-2)} = \frac{2}{2}=1$, $m_{EF} = \frac{5-3}{2-0}=1$ → collinear. 3. Find $k$ if line through $G(2, k)$ and $H(5, 8)$ has gradient $2$. Correction: $2 = \frac{8 - k}{5 - 2} \Rightarrow 6 = 8 - k \Rightarrow k = 2$. ## IV. Equation of a Straight Line ### Forms of equation A straight line can be expressed in multiple algebraic forms depending on given data. > **Slope-intercept form**: $y = mx + c$, where $m$ is gradient, $c$ is $y$-intercept. > **Point-slope form**: $y - y_1 = m(x - x_1)$, for line through $(x_1, y_1)$ with gradient $m$. > **Two-point form**: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$, for line through $(x_1, y_1)$ and $(x_2, y_2)$. > **General form**: $ax + by + c = 0$, where $a, b, c$ are constants, $a$ and $b$ not both zero. ### Properties - The $y$-intercept $c$ is found by setting $x=0$. - The $x$-intercept is found by setting $y=0$. - Converting between forms requires algebraic rearrangement. ### Solved example 1 Find equation of line with gradient $3$ passing through $(2, -1)$. Using point-slope: $$ y - (-1) = 3(x - 2) \Rightarrow y + 1 = 3x - 6 \Rightarrow y = 3x - 7 $$ Answer: $y = 3x - 7$. ### Solved example 2 Find equation of line through $A(-1, 4)$ and $B(3, -2)$. First, gradient: $m = \frac{-2 - 4}{3 - (-1)} = \frac{-6}{4} = -\frac{3}{2}$. Using point-slope with $A$: $$ y - 4 = -\frac{3}{2}(x - (-1)) \Rightarrow y - 4 = -\frac{3}{2}(x + 1) $$ Multiply by 2: $2y - 8 = -3x - 3 \Rightarrow 3x + 2y - 5 = 0$. Answer: $3x + 2y - 5 = 0$. ### Application exercises (corrected) 1. Equation of line with gradient $-\frac{1}{2}$ and $y$-intercept $4$. Correction: $y = -\frac{1}{2}x + 4$. 2. Equation of line through $(0, 0)$ and $(4, -6)$. Correction: $m = \frac{-6-0}{4-0} = -\frac{3}{2}$, so $y = -\frac{3}{2}x$ or $3x + 2y = 0$. 3. Convert $2x - 3y + 6 = 0$ to slope-intercept form. Correction: $-3y = -2x - 6 \Rightarrow y = \frac{2}{3}x + 2$. ## V. Intersecting Lines and Angle Between Two Lines ### Intersecting lines Two non-parallel lines intersect at exactly one point. To find the intersection, solve their equations simultaneously. ### Angle between two lines The acute angle $\theta$ between two lines with gradients $m_1$ and $m_2$ is given by: > $$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$$ > If $1 + m_1 m_2 = 0$, the lines are perpendicular ($\theta = 90^\circ$). ### Properties - The formula gives the acute angle; for obtuse, subtract from $180^\circ$. - If $m_1 = m_2$, lines are parallel ($\theta = 0^\circ$). - The denominator $1 + m_1 m_2$ must not be zero for the formula to apply (unless checking perpendicularity). ### Solved example 1 Find intersection of $L_1: 2x + y = 5$ and $L_2: x - y = 1$. Solve: From $L_2$, $x = y + 1$. Substitute into $L_1$: $2(y+1) + y = 5 \Rightarrow 2y + 2 + y = 5 \Rightarrow 3y = 3 \Rightarrow y = 1$, then $x = 2$. Answer: $(2, 1)$. ### Solved example 2 Find acute angle between lines with gradients $m_1 = 2$ and $m_2 = -\frac{1}{3}$. $$ \tan \theta = \left| \frac{-\frac{1}{3} - 2}{1 + (2)(-\frac{1}{3})} \right| = \left| \frac{-\frac{7}{3}}{1 - \frac{2}{3}} \right| = \left| \frac{-\frac{7}{3}}{\frac{1}{3}} \right| = | -7 | = 7 $$ $\theta = \tan^{-1}(7) \approx 81.87^\circ$. Answer: $\theta \approx 81.9^\circ$. ### Application exercises (corrected) 1. Intersection of $y = 3x - 2$ and $y = -x + 6$. Correction: $3x - 2 = -x + 6 \Rightarrow 4x = 8 \Rightarrow x=2$, $y=4$ → $(2,4)$. 2. Angle between lines $y = x + 1$ and $y = -2x + 3$. Correction: $m_1=1$, $m_2=-2$, $\tan \theta = \left| \frac{-2-1}{1+(1)(-2)} \right| = \left| \frac{-3}{-1} \right| = 3$ → $\theta \approx 71.6^\circ$. 3. Are lines $4x - 2y = 8$ and $2x - y = 3$ parallel? Correction: First: $y = 2x - 4$ ($m=2$); second: $y = 2x - 3$ ($m=2$) → parallel. ## VI. Parallel and perpendicular lines ### Conditions > Two lines are **parallel** if their gradients are equal: $m_1 = m_2$. > Two lines are **perpendicular** if the product of their gradients is $-1$: $m_1 \cdot m_2 = -1$, provided neither is vertical. ### Properties - A vertical line ($x = k$) is parallel to another vertical line; perpendicular to a horizontal line ($y = c$). - A horizontal line ($y = c$) has gradient $0$; perpendicular to vertical lines. - If $m_1 m_2 = -1$, then $m_2 = -\frac{1}{m_1}$ (negative reciprocal). ### Solved example 1 Find equation of line parallel to $y = -4x + 7$ passing through $(1, 2)$. Parallel → same gradient $m = -4$. Using point-slope: $y - 2 = -4(x - 1) \Rightarrow y - 2 = -4x + 4 \Rightarrow y = -4x + 6$. Answer: $y = -4x + 6$. ### Solved example 2 Find equation of line perpendicular to $2x + 3y = 6$ passing through $(-1, 4)$. First, rewrite: $3y = -2x + 6 \Rightarrow y = -\frac{2}{3}x + 2$ → $m_1 = -\frac{2}{3}$. Perpendicular gradient: $m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}$. Using point-slope: $y - 4 = \frac{3}{2}(x - (-1)) \Rightarrow y - 4 = \frac{3}{2}(x + 1)$. Multiply by 2: $2y - 8 = 3x + 3 \Rightarrow 3x - 2y + 11 = 0$. Answer: $3x - 2y + 11 = 0$. ### Application exercises (corrected) 1. Line parallel to $x - 2y = 5$ through $(3, -1)$. Correction: Rewrite: $y = \frac{1}{2}x - \frac{5}{2}$ → $m=\frac{1}{2}$. $y - (-1) = \frac{1}{2}(x - 3) \Rightarrow y+1 = \frac{1}{2}x - \frac{3}{2} \Rightarrow y = \frac{1}{2}x - \frac{5}{2}$. 2. Line perpendicular to $y = 5$ through $(-2, 3)$. Correction: $y=5$ is horizontal ($m=0$), so perpendicular is vertical → $x = -2$. 3. Are lines $y = -3x + 1$ and $6x + 2y = 10$ perpendicular? Correction: Second: $2y = -6x + 10 \Rightarrow y = -3x + 5$ → $m_1 = m_2 = -3$ → parallel, not perpendicular. ## VII. Distance of a point from a line ### Definition and formula The perpendicular distance from a point to a line is the shortest distance. > The distance $d$ from point $P(x_0, y_0)$ to line $ax + by + c = 0$ is: > $$d = \frac{\vert ax_0 + by_0 + c \vert}{\sqrt{a^2 + b^2}}$$ ### Properties - Distance is always non-negative. - If $d = 0$, the point lies on the line. - The formula derives from projecting the point onto the line along the normal vector. ### Solved example 1 Find distance from $P(3, -1)$ to line $2x - 3y + 6 = 0$. $$ d = \frac{\vert 2(3) - 3(-1) + 6 \vert}{\sqrt{2^2 + (-3)^2}} = \frac{\vert 6 + 3 + 6 \vert}{\sqrt{4+9}} = \frac{15}{\sqrt{13}} = \frac{15\sqrt{13}}{13} $$ Answer: $\frac{15\sqrt{13}}{13}$ units. ### Solved example 2 Find distance from $Q(-2, 4)$ to line $x = 5$. Rewrite line: $x - 5 = 0$ → $a=1, b=0, c=-5$. $$ d = \frac{\vert 1(-2) + 0(4) - 5 \vert}{\sqrt{1^2 + 0^2}} = \frac{\vert -2 -5 \vert}{1} = 7 $$ Answer: $7$ units. ### Application exercises (corrected) 1. Distance from $R(0, 0)$ to line $4x + 3y - 12 = 0$. Correction: $d = \frac{\vert 0+0-12 \vert}{\sqrt{16+9}} = \frac{12}{5} = 2.4$. 2. Distance from $S(1, -2)$ to line $y = 2x + 1$. Correction: Rewrite: $2x - y + 1 = 0$ → $d = \frac{\vert 2(1) - (-2) + 1 \vert}{\sqrt{4+1}} = \frac{\vert 2+2+1 \vert}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$. 3. Distance from $T(-3, 5)$ to line $x = -1$. Correction: Line: $x + 1 = 0$ → $d = \frac{\vert -3 + 1 \vert}{\sqrt{1}} = \vert -2 \vert = 2$. ## VIII. Reduction to linear form and draw straight line graphs for the linear forms 1 ### Concept Non-linear relationships can be transformed into linear form $Y = mX + c$ by suitable substitution, enabling graphing as a straight line and determination of constants. > To reduce to linear form, identify variables $X$ and $Y$ (functions of original variables) such that the equation becomes $Y = mX + c$. ### Properties - After substitution, plotting $Y$ vs $X$ yields a straight line. - Gradient $m$ and intercept $c$ can be found from the graph. - Common forms: $y = ax^n \rightarrow \log y = \log a + n \log x$; $y = ae^{bx} \rightarrow \ln y = \ln a + bx$. ### Solved example 1 Given $y = ax^b$, reduce to linear form and state what to plot. Take natural log: $\ln y = \ln a + b \ln x$. Let $Y = \ln y$, $X = \ln x$, then $Y = bX + \ln a$. Plot $\ln y$ against $\ln x$ → straight line with gradient $b$, intercept $\ln a$. ### Solved example 2 For $y = \frac{k}{x} + c$, reduce to linear form. Rewrite: $y = k \cdot \frac{1}{x} + c$. Let $Y = y$, $X = \frac{1}{x}$, then $Y = kX + c$. Plot $y$ against $\frac{1}{x}$ → straight line with gradient $k$, intercept $c$. ### Application exercises (corrected) 1. Reduce $y = a \cdot 2^{bx}$ to linear form. Correction: $\ln y = \ln a + bx \ln 2$ → let $Y = \ln y$, $X = x$, then $Y = (b \ln 2)X + \ln a$. 2. Given $y = \frac{a}{x^2} + b$, state what to plot. Correction: Let $Y = y$, $X = \frac{1}{x^2}$, then $Y = aX + b$ → plot $y$ vs $\frac{1}{x^2}$. 3. If a plot of $\log y$ vs $x$ is linear, what is the original relationship? Correction: $\log y = mx + c \Rightarrow y = 10^{mx + c} = 10^c \cdot (10^m)^x$ → $y = k \cdot a^x$ (exponential). ## IX. Reduction to linear form and draw straight line graphs for the linear forms 2 ### Additional forms This section covers further transformations for power, exponential, and rational functions. > For $y = ax^b$, plot $\log y$ vs $\log x$ → gradient $b$, intercept $\log a$. > For $y = ae^{bx}$, plot $\ln y$ vs $x$ → gradient $b$, intercept $\ln a$. > For $y = \frac{a}{x} + b$, plot $y$ vs $\frac{1}{x}$ → gradient $a$, intercept $b$. ### Properties - The choice of transformation depends on the suspected form. - Logarithms (base 10 or natural) are used for power/exponential laws. - Intercepts and gradients yield physical constants in applied contexts. ### Solved example 1 Experimental data suggests $y = kx^n$. State how to find $k$ and $n$ graphically. Plot $\log_{10} y$ against $\log_{10} x$. Gradient = $n$, intercept = $\log_{10} k$ → $k = 10^{\text{intercept}}$. ### Solved example 2 Given $y = \frac{a}{x^2} + b$, and a plot of $y$ vs $\frac{1}{x^2}$ gives a line with gradient $4$ and intercept $-3$. Then $a = 4$, $b = -3$ → original equation: $y = \frac{4}{x^2} - 3$. ### Application exercises (corrected) 1. A graph of $\ln y$ vs $x$ is linear with gradient $-0.5$ and intercept $2$. Find original equation. Correction: $\ln y = -0.5x + 2 \Rightarrow y = e^{2} \cdot e^{-0.5x} \Rightarrow y = e^2 e^{-0.5x}$. 2. For $y = ax^b$, if plot of $\log y$ vs $\log x$ has gradient $1.5$ and intercept $\log 3$, find $y$ when $x=10$. Correction: $b=1.5$, $a=3$ → $y = 3 \cdot (10)^{1.5} = 3 \cdot 31.62 \approx 94.86$. 3. Reduce $T = 2\pi \sqrt{\frac{l}{g}}$ to linear form to find $g$. Correction: Square: $T^2 = 4\pi^2 \frac{l}{g} \Rightarrow T^2 = \left( \frac{4\pi^2}{g} \right) l$ → plot $T^2$ vs $l$ → gradient $= \frac{4\pi^2}{g}$ → $g = \frac{4\pi^2}{\text{gradient}}$. ## X. Loci using Cartesian or parametric forms 1 ### Definition A locus is a set of points satisfying a given condition. > In Cartesian form, a locus is an equation in $x$ and $y$ whose solution set is the locus. > In parametric form, coordinates are expressed as functions of a parameter $t$: $x = f(t)$, $y = g(t)$. ### Properties - To find Cartesian locus: express condition algebraically, simplify to equation in $x,y$. - To find parametric locus: eliminate parameter $t$ to get Cartesian equation, or state parametric equations directly. - Common loci: circle, parabola, ellipse, hyperbola, straight line. ### Solved example 1 Find locus of point $P(x,y)$ equidistant from $A(0,0)$ and $B(4,0)$. Condition: $PA = PB$. $$ \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{(x-4)^2 + (y-0)^2} $$ Square both sides: $x^2 + y^2 = (x-4)^2 + y^2 \Rightarrow x^2 = x^2 - 8x + 16 \Rightarrow 8x = 16 \Rightarrow x = 2$. Answer: $x = 2$ (vertical line). ### Solved example 2 Find locus of point $P$ such that its distance from $(0,2)$ is twice its distance from line $y = -1$. Let $P(x,y)$. Distance to point: $\sqrt{x^2 + (y-2)^2}$. Distance to line $y = -1$: $\vert y - (-1) \vert = \vert y+1 \vert$. Condition: $\sqrt{x^2 + (y-2)^2} = 2 \vert y+1 \vert$. Square: $x^2 + (y-2)^2 = 4(y+1)^2$. Expand: $x^2 + y^2 - 4y + 4 = 4(y^2 + 2y + 1) = 4y^2 + 8y + 4$. Simplify: $x^2 + y^2 - 4y + 4 - 4y^2 - 8y - 4 = 0 \Rightarrow x^2 - 3y^2 - 12y = 0$. Answer: $x^2 - 3y^2 - 12y = 0$. ### Application exercises (corrected) 1. Locus of point $P$ such that $PA^2 + PB^2 = 10$, where $A(-1,0)$, $B(1,0)$. Correction: $PA^2 = (x+1)^2 + y^2$, $PB^2 = (x-1)^2 + y^2$ Sum: $(x^2+2x+1+y^2) + (x^2-2x+1+y^2) = 2x^2 + 2y^2 + 2 = 10$ $\Rightarrow 2x^2 + 2y^2 = 8 \Rightarrow x^2 + y^2 = 4$ → circle centre $(0,0)$, radius $2$. 2. Locus of point $P(x,y)$ such that slope of line joining $P$ to $(2,3)$ is $-1$. Correction: $\frac{y-3}{x-2} = -1 \Rightarrow y-3 = -x+2 \Rightarrow y = -x + 5$. 3. Parametric equations: $x = 2\cos t$, $y = 3\sin t$. Find Cartesian locus. Correction: $\frac{x}{2} = \cos t$, $\frac{y}{3} = \sin t$ → $\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = \cos^2 t + \sin^2 t = 1$ → $\frac{x^2}{4} + \frac{y^2}{9} = 1$ (ellipse). ## Synthèse du chapitre - **Distance and midpoint**: Use $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and $\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$ for length and centre of segments. - **Section formula**: Internal: $\left( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n} \right)$; External: $\left( \frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n} \right)$ for points dividing in ratio $m:n$. - **Gradient**: $m = \frac{y_2-y_1}{x_2-x_1}$ determines slope; equal gradients → parallel; product $-1$ → perpendicular. - **Line equations**: Slope-intercept ($y=mx+c$), point-slope ($y-y_1=m(x-x_1)$), two-point, general ($ax+by+c=0$). - **Intersection and angle**: Solve equations simultaneously for intersection; $\tan \theta = \left| \frac{m_2-m_1}{1+m_1m_2} \right|$ for acute angle. - **Point-to-line distance**: $d = \frac{\vert ax_0+by_0+c \vert}{\sqrt{a^2+b^2}}$ from point to line $ax+by+c=0$. - **Linearisation**: Transform $y=ax^b$ → $\log y = \log a + b \log x$; $y=ae^{bx}$ → $\ln y = \ln a + bx$; $y=\frac{a}{x}+c$ → $y$ vs $\frac{1}{x}$ linear. - **Loci**: Translate geometric conditions into algebraic equations (Cartesian) or $x=f(t), y=g(t)$ (parametric); eliminate parameter to verify. ## Glossaire - **Cartesian plane**: Coordinate system with perpendicular $x$ and $y$ axes. - **Gradient**: Measure of steepness of a line, ratio of vertical to horizontal change. - **Locus**: Set of points satisfying a given geometric condition. - **Midpoint**: Point equidistant from the endpoints of a segment. - **Parallel lines**: Lines in a plane that never intersect; have equal gradients. - **Perpendicular lines**: Lines intersecting at $90^\circ$; product of gradients is $-1$. - **Section formula**: Gives coordinates of point dividing a segment in a given ratio internally or externally. - **Linear form**: Equation of type $Y = mX + c$ after variable substitution. - **Parametric equations**: Coordinates expressed as functions of a third variable (parameter). - **Point-to-line distance**: Shortest distance from a point to a line, measured perpendicularly. - **Slope-intercept form**: $y = mx + c$, where $m$ is gradient and $c$ is $y$-intercept. - **Two-point form**: Equation of line through two points using their coordinates.