COORDINATE GEOMETRY · HL
The Line
Plot it, measure it, find its equation.
Section 1 of 6
Plotting points
Every point on a Cartesian plane has two coordinates: $(x,\,y)$.
The $x$ comes first — across the page. The $y$ comes second — up the page. Across, then up. Always in that order.
The two lines that cross at $(0,\,0)$ are the axes. The horizontal one is the $x$-axis, the vertical one is the $y$-axis.
(i) Worked example
On a Cartesian plane plot the places Shop $(1,\,3)$, Pub $(-2,\,-3)$ and Church $(1,\,-4)$.
Shop: from the origin go across $1$, then up $3$.
Pub: across $-2$ (i.e. left $2$), then down $3$.
Church: across $1$, then down $4$.
YOU TRY · 1
Plot the points $A(3,\,2)$, $B(-1,\,4)$ and $C(-2,\,-3)$ on the same Cartesian plane.
Across first, then up. Mind the minus signs.
$A$: across $3$, up $2$. $B$: left $1$, up $4$. $C$: left $2$, down $3$.
Three dots, correctly placed.
YOU TRY · 2
Which quadrant does the point $(-4,\,-1)$ lie in?
Both coordinates negative — left and down.
$x < 0$ and $y < 0$ ⇒ bottom-left.
The third quadrant.
YOU TRY · 3
Write down the coordinates of the point on the $x$-axis that is $5$ units to the left of the origin.
On the $x$-axis, the $y$-coordinate is always $0$.
Left $5$ ⇒ $x = -5$, and on the axis $y = 0$.
$(-5,\,0)$
Section 2 of 6
Lines in real life
A straight line can model money over time. Build a table, plot it, read it off.
(i) Worked example — income & expenses
Income in a shop is €$30$ per hour. Expenses are wages €$20$ per hour and €$50$ rent. Show income and expenses on two tables and on one diagram.
Income rises by €$30$ every hour — nothing to start with. Expenses start at €$50$ (the rent) and rise by €$20$ every hour.
| Income | |
| Time | Money |
| 0 | 0 |
| 1 | 30 |
| 2 | 60 |
| 3 | 90 |
| $x$ | $y = 30x$ |
| Expenses | |
| Time | Money |
| 0 | 50 |
| 1 | 70 |
| 2 | 90 |
| 3 | 110 |
| $x$ | $y = 20x + 50$ |
(ii) Where do they meet?
The lines cross where income equals expenses. Set them equal:
$30x = 20x + 50$
$10x = 50$
$x = 5$
After $5$ hours the shop breaks even — income catches up with expenses.
YOU TRY · 4
A taxi charges €$4$ call-out plus €$2$ per km. Write the cost $y$ as an equation in the distance $x$ (km).
Fixed charge first, then the bit that grows with $x$.
Start at €$4$; add €$2$ for each km.
$y = 2x + 4$
YOU TRY · 5
Two plans cost $y = 15 + 0.5x$ and $y = 25$. After how many units $x$ do they cost the same?
Set them equal and solve for $x$.
$15 + 0.5x = 25$
$0.5x = 10$
$x = 20$
YOU TRY · 6
The lines $y = 4x$ and $y = x + 9$ meet. Find the value of $x$ where they cross.
Equal $y$ values means equal right-hand sides.
$4x = x + 9$
$3x = 9$
$x = 3$
Section 3 of 6
Distance between two points
Distance is the measure $|AB|$ — the length of the segment from $A$ to $B$.
Drop a right-angled triangle under the segment. The horizontal leg is $x_2 - x_1$, the vertical leg is $y_2 - y_1$, and $|AB|$ is the hypotenuse.
By Pythagoras:
Must learn
1.$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
(i) Worked example
$A(-3,\,1)$ and $B(5,\,-2)$. Find $|AB|$.
Label them: $(\underset{x_1}{-3},\,\underset{y_1}{1}) \quad (\underset{x_2}{5},\,\underset{y_2}{-2})$
$|AB| = \sqrt{(5 - (-3))^2 + (-2 - 1)^2}$
$= \sqrt{(8)^2 + (-3)^2} = \sqrt{64 + 9}$
$|AB| = \sqrt{73}$
(ii) Working backwards for a coordinate
$P(5,\,-1)$ and $Q(4,\,k)$ are two points such that $|PQ| = \sqrt{5}$. Find the two values of $k$.
$\sqrt{(4 - 5)^2 + (k - (-1))^2} = \sqrt{5}$
$1 + (k + 1)^2 = 5$
$(k + 1)^2 = 4$
$k + 1 = 2 \quad$ or $\quad k + 1 = -2$
$k = 1 \quad$ or $\quad k = -3$
Squaring loses the sign, so a length condition usually gives two answers. Keep both.
YOU TRY · 7
Find $|AB|$ where $A(2,\,1)$ and $B(5,\,5)$.
Differences first, then square, add, root.
$\sqrt{(5-2)^2 + (5-1)^2} = \sqrt{9 + 16} = \sqrt{25}$
$|AB| = 5$
YOU TRY · 8
Find the distance between $(-1,\,-2)$ and $(3,\,1)$.
Mind the double-negatives: $3 - (-1) = 4$.
$\sqrt{(3+1)^2 + (1+2)^2} = \sqrt{16 + 9} = \sqrt{25}$
$5$
YOU TRY · 9
$P(2,\,3)$ and $Q(2,\,k)$ with $|PQ| = 4$. Find the two values of $k$.
Same $x$ both points, so the first bracket is $0$.
$(k - 3)^2 = 16$
$k - 3 = 4$ or $k - 3 = -4$
$k = 7 \quad$ or $\quad k = -1$
Section 4 of 6
Midpoint
The midpoint is the point exactly halfway along the segment — the average of the two coordinates.
Must learn
1.Midpoint $= \left(\dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2}\right)$
(i) Worked example
$A(3,\,1)$ and $B(-5,\,6)$. Find the midpoint.
Midpoint $= \left(\dfrac{3 + (-5)}{2},\ \dfrac{1 + 6}{2}\right)$
Midpoint $= \left(-1,\ \dfrac{7}{2}\right)$
(ii) Working backwards for an endpoint
$A(5,\,-1)$ is the midpoint of $P(1,\,3)$ and $Q$. Find $Q$.
Two ways. Either set up the equations, or "jump the same distance again."
$x$: $\dfrac{1 + x}{2} = 5 \ \Rightarrow\ 1 + x = 10 \ \Rightarrow\ x = 9$
$y$: $\dfrac{3 + y}{2} = -1 \ \Rightarrow\ 3 + y = -2 \ \Rightarrow\ y = -5$
Jump check: $\ 1 \to 5 \to 9 \ $ (add $4$ twice), $\ 3 \to -1 \to -5 \ $ (subtract $4$ twice).
$Q = (9,\,-5)$
YOU TRY · 10
Find the midpoint of $A(2,\,6)$ and $B(8,\,2)$.
Average the $x$'s, average the $y$'s.
$\left(\dfrac{2+8}{2},\ \dfrac{6+2}{2}\right)$
$(5,\,4)$
YOU TRY · 11
Find the midpoint of $(-3,\,4)$ and $(5,\,-2)$.
Negatives add in too — take your time.
$\left(\dfrac{-3+5}{2},\ \dfrac{4-2}{2}\right)$
$(1,\,1)$
YOU TRY · 12
$M(4,\,3)$ is the midpoint of $A(1,\,7)$ and $B$. Find $B$.
Jump from $A$ to $M$, then the same jump again.
$1 \to 4 \to 7$ (add $3$), $7 \to 3 \to -1$ (subtract $4$).
$B = (7,\,-1)$
Section 5 of 6
Slope
Slope is the rate of change — how steep the line is. Same idea as $\dfrac{dy}{dx}$.
Slope is rise over run — and that is exactly $\tan\theta$:
$m = \dfrac{\text{rise}}{\text{run}} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{\text{change in } y}{\text{change in } x}$
Must learn
1.$m = \dfrac{y_2 - y_1}{x_2 - x_1}$
2.$m = \tan\theta = \dfrac{dy}{dx}$ (rate of change of $y$ with respect to $x$)
(i) Worked example
Find the slope of the line through $(1,\,5)$ and $(-2,\,9)$.
$m = \dfrac{9 - 5}{-2 - 1} = \dfrac{4}{-3}$
$m = -\dfrac{4}{3}$
(ii) Drawing a slope
A slope is just over then up (or down). Read the fraction as run-over-rise.
Slope $\dfrac{2}{3}$: over $3$, then up $2$.
Slope $-\dfrac{3}{2}$: over $2$, then down $3$.
YOU TRY · 13
Find the slope of the line through $(2,\,1)$ and $(6,\,9)$.
$\dfrac{y_2 - y_1}{x_2 - x_1}$ — keep the order the same top and bottom.
$m = \dfrac{9 - 1}{6 - 2} = \dfrac{8}{4}$
$m = 2$
YOU TRY · 14
Find the slope of the line through $(3,\,5)$ and $(-1,\,5)$.
What is the change in $y$ here?
$m = \dfrac{5 - 5}{-1 - 3} = \dfrac{0}{-4}$
$m = 0$ — a flat, horizontal line.
YOU TRY · 15
On graph paper, draw a line with slope $\dfrac{3}{4}$.
Bottom number is the "over", top number is the "up".
From any point: over $4$, then up $3$. Join.
A line rising over $4$, up $3$.
Section 6 of 6
Equation of a line
Take a fixed point $(x_1,\,y_1)$ on the line and any general point $(x,\,y)$. The slope between them is $m$:
$m = \dfrac{y - y_1}{x - x_1}$
Cross-multiply and you have the line. Learn it in this form:
Must learn
1.$y - y_1 = m(x - x_1)$
2.Tidy into the form $ax + by = c$.
3.$m = \tan\theta$ — slope from an angle to the $x$-axis.
(i) Point and slope
Find the line through $(1,\,-3)$ with slope $-\dfrac{5}{6}$.
$y - y_1 = m(x - x_1)$
$y + 3 = -\dfrac{5}{6}(x - 1)$
$6(y + 3) = -5(x - 1)$
$6y + 18 = -5x + 5$
$5x + 6y = -13$
(ii) Two points
Find the line through $(3,\,-5)$ and $(-1,\,4)$.
$m = \dfrac{4 - (-5)}{-1 - 3} = \dfrac{9}{-4} = -\dfrac{9}{4}$
$y + 5 = -\dfrac{9}{4}(x - 3)$
$4y + 20 = -9x + 27$
$9x + 4y = 7$
(iii) Equal intercepts
Find the equations of the lines through $(1,\,5)$ which cut equal intercepts with both axes.
Equal intercepts on the axes means the line makes a $45^\circ$ triangle with them. So $m = \tan 45^\circ = 1$ or $m = \tan 135^\circ = -1$. Two answers.
Line 1: $m = 1 \ \Rightarrow\ y - 5 = 1(x - 1)$
$x - y = -4 \quad$ (i.e. $y = x + 4$)
Line 2: $m = -1 \ \Rightarrow\ y - 5 = -1(x - 1)$
$x + y = 6$
YOU TRY · 16
Find the equation of the line through $(2,\,-1)$ with slope $3$. Give it in the form $ax + by = c$.
$y - y_1 = m(x - x_1)$, then tidy up.
$y + 1 = 3(x - 2)$
$y + 1 = 3x - 6$
$3x - y = 7$
YOU TRY · 17
Find the equation of the line through $(0,\,4)$ and $(2,\,0)$.
Slope first, then point-slope with either point.
$m = \dfrac{0 - 4}{2 - 0} = -2$
$y - 4 = -2(x - 0)$
$2x + y = 4$
YOU TRY · 18
Find the equation of the line through $(2,\,3)$ that makes a $45^\circ$ angle with the positive $x$-axis.
Angle to the $x$-axis means $m = \tan\theta$.
$m = \tan 45^\circ = 1$
$y - 3 = 1(x - 2)$
$x - y = -1 \quad$ (i.e. $y = x + 1$)
SUM
The lot in one box
The Line toolkit
1.Plot $(x,\,y)$: across first, then up.
2.Distance $|AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
3.Midpoint $= \left(\dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2}\right)$
4.Slope $m = \dfrac{y_2 - y_1}{x_2 - x_1} = \tan\theta = \dfrac{dy}{dx}$
5.Equation: $y - y_1 = m(x - x_1)$, then tidy to $ax + by = c$.
6.A length condition gives two answers — keep both.
7.Equal intercepts ⇒ $m = \tan 45^\circ = 1$ or $m = \tan 135^\circ = -1$.
End of lesson
The Line — HL · Mathslive.ie