Co-ordinate Geometry · Junior Cert
The Line
Junior Cert Higher · Midpoint, distance & slope · Tap NEXT to begin
Section 1 of 5
The Cartesian Plane
Every point is fixed by two numbers, its co-ordinates $(x,\,y)$. The grid is the Cartesian plane; the lines are the $x$-axis and $y$-axis; they cross at the origin $(0,0)$.
Reading co-ordinates
1$x$ first: left is $-$, right is $+$.
2$y$ second: down is $-$, up is $+$.
3Origin $=(0,0)$, where the axes meet.
You try
Which point sits on the $y$-axis: $A(-3,1)$, $C(0,4)$ or $D(-1,0)$?
A point on the $y$-axis has $x=0$.
On the $y$-axis the $x$ co-ordinate is $0$.
$C(0,4)$ has $x=0$.
$C(0,4)$
Section 2 of 5
The Midpoint
The midpoint is the middle of $[AB]$ — average the $x$’s and average the $y$’s.
Midpoint formula (in your tables)
1$\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$
Worked: midpoint of $A(-2,-1)$ and $B(3,2)$ $=\left(\dfrac{-2+3}{2},\dfrac{-1+2}{2}\right)=\left(\dfrac{1}{2},\dfrac{1}{2}\right)$.
You try
Find the midpoint of $P(4,7)$ and $Q(10,1)$.
Average the $x$’s, then the $y$’s.
$\left(\dfrac{4+10}{2},\dfrac{7+1}{2}\right)$
$=\left(\dfrac{14}{2},\dfrac{8}{2}\right)$
$(7,\,4)$
Working backwards: if you know the midpoint and one end, use the same jumps to reach the other end (central symmetry).
You try
$P(1,-2)$ is the midpoint of $A(-1,4)$ and $B$. Find $B$.
Find the jump from $A$ to $P$, then repeat it.
$A \to P$: $x$: $-1 \to 1$ ($+2$); $y$: $4 \to -2$ ($-6$).
Apply again from $P$: $x:\,1+2=3$; $y:\,-2-6=-8$.
$B(3,\,-8)$
Section 3 of 5
Distance $|AB|$
The distance between two points is the length of $[AB]$. Drop a right-angled triangle under the segment and use Pythagoras: the horizontal run and vertical rise are the two legs, $|AB|$ is the hypotenuse.
Distance formula (in your tables)
1$|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Worked: $A(-2,-1)$, $B(3,2)$: run $=5$, rise $=3$, so $|AB|=\sqrt{5^2+3^2}=\sqrt{34}$.
You try
Find $|PQ|$ for $P(3,1)$ and $Q(4,7)$.
Subtract the co-ordinates, square, add, root.
$\sqrt{(4-3)^2+(7-1)^2}$
$=\sqrt{1^2+6^2}=\sqrt{1+36}$
$|PQ|=\sqrt{37}$
Section 4 of 5
Slope
Slope measures steepness: how much the line rises for how much it runs.
Slope $m$
1$m=\dfrac{\text{rise}}{\text{run}}=\dfrac{y_2-y_1}{x_2-x_1}$
2Up to the right $\Rightarrow m>0$; down to the right $\Rightarrow m<0$.
Worked: $A(-3,2)$, $B(2,-3)$: $m=\dfrac{-3-2}{2-(-3)}=\dfrac{-5}{5}=-1$.
You try
Find the slope of the line through $A(1,2)$ and $B(4,8)$.
$m=\dfrac{y_2-y_1}{x_2-x_1}$.
$m=\dfrac{8-2}{4-1}=\dfrac{6}{3}$
$m=2$
Section 5 of 5
Putting It Together
A full exam part often asks for all three at once. Plot the points, then apply each formula.
You try
For $A(-3,2)$ and $B(2,-3)$ find (i) the midpoint, (ii) $|AB|$, (iii) the slope of $AB$.
Midpoint = average; distance = Pythagoras; slope = rise over run.
(i) $\left(\dfrac{-3+2}{2},\dfrac{2-3}{2}\right)=\left(-\dfrac12,-\dfrac12\right)$
(ii) $\sqrt{(2+3)^2+(-3-2)^2}=\sqrt{25+25}=\sqrt{50}$
(iii) $m=\dfrac{-3-2}{2+3}=\dfrac{-5}{5}=-1$
midpoint $\left(-\tfrac12,-\tfrac12\right)$, $|AB|=\sqrt{50}$, $m=-1$
More examples
Worked example — plotting points
Plot $A(3,2)$ and $B(-1,-4)$ on the plane, and state which quadrant each is in.
$A(3,2)$: right $3$, up $2$ → quadrant $1$
$B(-1,-4)$: left $1$, down $4$ → quadrant $3$
$A$ in Q1, $B$ in Q3
You try
Find the midpoint of $(3,5)$ and $(7,9)$.
Average the $x$’s and the $y$’s.
$\left(\dfrac{3+7}{2},\dfrac{5+9}{2}\right)$
$(5,\,7)$
That’s The Line — the three core tools.
Midpoint, distance and slope are the foundation. Next in this topic: the equation of a line, $y-y_1=m(x-x_1)$.