Coordinate geometry: points, midpoints and distances

Coordinate geometry turns a diagram into algebra. A point is described by an ordered pair, and lengths and positions can be calculated exactly rather than estimated from a drawing.

These foundations support straight lines, circles, vectors, calculus and mechanics. The formulae are simple, but signs, subtraction order and exact values need care.

Before you begin

You should be able to:

• work confidently with negative numbers;
• substitute into formulae;
• square integers and simplify square roots;
• use Pythagoras’ theorem.

Review exact arithmetic, indices, roots and surds or Pythagoras and trigonometry if needed.

Coordinates are ordered pairs

A point $P$ in the Cartesian plane has coordinates

$$P(x,y).$$

The first coordinate gives horizontal position and the second gives vertical position. Starting at the origin $O(0,0)$:

• positive $x$ means move right;
• negative $x$ means move left;
• positive $y$ means move up;
• negative $y$ means move down.

Order matters. The points $(2,-5)$ and $(-5,2)$ are different.

The axes split the plane into four quadrants:

Quadrant	Sign of $x$	Sign of $y$
I	$+$	$+$
II	$-$	$+$
III	$-$	$-$
IV	$+$	$-$

A point on an axis is not in any quadrant. For example, $(0,4)$ is on the $y$ axis and $(-3,0)$ is on the $x$ axis.

Worked example 1: reflect a point

Let $A(-3,5)$.

1. Reflection in the $x$ axis changes the sign of the $y$ coordinate: $(-3,5)\mapsto(-3,-5).$
2. Reflection in the $y$ axis changes the sign of the $x$ coordinate: $(-3,5)\mapsto(3,5).$
3. Reflection in the origin changes both signs: $(-3,5)\mapsto(3,-5).$
4. Reflection in the line $y=x$ swaps the coordinates: $(-3,5)\mapsto(5,-3).$

These rules follow from geometry. They are not all versions of “change the signs”.

Horizontal and vertical change

Suppose

$$A(x_1,y_1),\qquad B(x_2,y_2).$$

Moving from $A$ to $B$ produces the signed changes

$$\boxed{\Delta x=x_2-x_1},\qquad \boxed{\Delta y=y_2-y_1}.$$

The symbol $\Delta$ means change in. A negative change is meaningful: it records leftward or downward movement.

Worked example 2: calculate a displacement

For $A(-4,3)$ and $B(2,-5)$,

$$\Delta x=2-(-4)=6,$$

and

$$\Delta y=-5-3=-8.$$

Thus the move from $A$ to $B$ is $6$ units right and $8$ units down. Reversing the journey reverses both signs:

$$B\text{ to }A:\qquad \Delta x=-6,\quad \Delta y=8.$$

This ordered change $(\Delta x,\Delta y)$ is also the vector from $A$ to $B$. See vectors for vector notation and operations.

The midpoint of a line segment

The midpoint lies halfway between the endpoints in both coordinate directions. For $A(x_1,y_1)$ and $B(x_2,y_2)$,

$$\boxed{M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)}.$$

This is simply the mean of the two $x$ coordinates and the mean of the two $y$ coordinates.

Worked example 3: find a midpoint

Find the midpoint of $A(-7,4)$ and $B(5,-2)$.

$$M=\left(\frac{-7+5}{2},\frac{4+(-2)}{2}\right) =\left(\frac{-2}{2},\frac{2}{2}\right) =(-1,1).$$

Check geometrically. From $A$ to $M$ the change is $(6,-3)$, and from $M$ to $B$ it is also $(6,-3)$. Therefore $M$ really is halfway along the segment.

Worked example 4: recover a missing endpoint

The midpoint of $A(3,-4)$ and $B(x,y)$ is $M(-2,5)$. Find $B$.

Use each coordinate separately:

$$\frac{3+x}{2}=-2 \quad\Longrightarrow\quad 3+x=-4 \quad\Longrightarrow\quad x=-7,$$ $$\frac{-4+y}{2}=5 \quad\Longrightarrow\quad -4+y=10 \quad\Longrightarrow\quad y=14.$$

Hence

$$\boxed{B(-7,14)}.$$

A quicker check is that $M-A=(-5,9)$, so $B=M+(-5,9)=(-7,14)$.

Common misconception: halve each coordinate

The midpoint of $(2,8)$ and $(10,4)$ is not $(1,4)$ or $(5,2)$. You must first add corresponding coordinates, then divide by $2$:

$$\left(\frac{2+10}{2},\frac{8+4}{2}\right)=(6,6).$$

Distance between two points

The horizontal and vertical changes between $A(x_1,y_1)$ and $B(x_2,y_2)$ form the perpendicular sides of a right angled triangle. By Pythagoras’ theorem,

$$AB^2=(x_2-x_1)^2+(y_2-y_1)^2.$$

Since distance is non-negative,

$$\boxed{AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}.$$

Subtraction order does not affect the distance because each difference is squared. It is nevertheless safest to use one consistent order.

Worked example 5: calculate an exact distance

Find the distance between $P(-2,7)$ and $Q(4,-1)$.

First find the coordinate changes:

$$\Delta x=4-(-2)=6,\qquad \Delta y=-1-7=-8.$$

Then

$$PQ=\sqrt{6^2+(-8)^2} =\sqrt{36+64} =\sqrt{100} =10.$$

The sign of $-8$ describes direction, but its contribution to length is $(-8)^2=64$.

Worked example 6: simplify a surd distance

Find the distance from $A(-1,2)$ to $B(5,7)$.

$$AB=\sqrt{(5-(-1))^2+(7-2)^2} =\sqrt{6^2+5^2} =\sqrt{61}.$$

Because $61$ has no square factor greater than $1$, $\sqrt{61}$ is the exact answer. A decimal approximation is

$$\sqrt{61}\approx7.81.$$

Unless a question requests a decimal or stated accuracy, keep the exact surd.

Special cases

If two points have the same $y$ coordinate, their separation is horizontal:

$$(x_1,c)\text{ to }(x_2,c):\qquad |x_2-x_1|.$$

If they have the same $x$ coordinate, their separation is vertical:

$$(c,y_1)\text{ to }(c,y_2):\qquad |y_2-y_1|.$$

The absolute value is necessary because a length cannot be negative.

Common misconception: adding coordinate changes

For a move $3$ units right and $4$ units up, the straight line distance is not $3+4=7$. That sum is the length of a route along the two perpendicular sides. The direct distance is

$$\sqrt{3^2+4^2}=5.$$

Dividing a segment in a given ratio

Midpoint is the special case of dividing a segment in the ratio $1:1$. More generally, suppose $P$ divides the segment from $A(x_1,y_1)$ to $B(x_2,y_2)$ internally so that

$$AP:PB=m:n.$$

Then $P$ is the fraction $\dfrac{m}{m+n}$ of the way from $A$ to $B$:

$$P=A+\frac{m}{m+n}(B-A).$$

In coordinates,

$$\boxed{P\left(\frac{nx_1+mx_2}{m+n},\frac{ny_1+my_2}{m+n}\right)}.$$

Notice the crossed weights. The coefficient of $A$ is $n$, while the coefficient of $B$ is $m$. The movement form is often easier to remember and less prone to error.

Worked example 7: divide a segment

Points $A(-2,1)$ and $B(10,7)$ are joined. Point $P$ satisfies $AP:PB=1:2$. Find $P$.

Because the whole segment contains $1+2=3$ equal parts, $P$ is one third of the way from $A$ to $B$.

$$B-A=(10-(-2),7-1)=(12,6).$$

One third of this change is $(4,2)$, so

$$P=(-2,1)+(4,2)=(2,3).$$

Check:

$$P-A=(4,2),\qquad B-P=(8,4)=2(4,2).$$

Therefore $AP:PB=1:2$, as required. A point in a $1:2$ ratio is closer to $A$, not closer to $B$.

Using coordinates to prove geometric facts

A diagram can suggest a result, but coordinate calculations can establish it. Useful tests include:

Claim	Coordinate evidence
Same point	both coordinates are equal
Same midpoint	midpoint formula gives identical coordinates
Equal lengths	squared distances are equal
Point lies on perpendicular bisector of $AB$	its distances from $A$ and $B$ are equal
Parallelogram	diagonals have the same midpoint
Isosceles triangle	two side lengths are equal

When comparing lengths, use squared distances. This avoids unnecessary square roots: for non-negative lengths, $AB=AC$ exactly when $AB^2=AC^2$.

Worked example 8: prove a triangle is isosceles

Let $A(-1,2)$, $B(5,4)$ and $C(3,-2)$. Show that triangle $ABC$ is isosceles.

Calculate squared side lengths:

$$AB^2=(5-(-1))^2+(4-2)^2=6^2+2^2=40,$$ $$AC^2=(3-(-1))^2+(-2-2)^2=4^2+(-4)^2=32,$$ $$BC^2=(3-5)^2+(-2-4)^2=(-2)^2+(-6)^2=40.$$

Thus $AB^2=BC^2$, so $AB=BC$. Therefore

$$\boxed{\triangle ABC\text{ is isosceles at }B.}$$

Naming the equal sides also identifies the vertex where they meet.

Worked example 9: prove a quadrilateral is a parallelogram

Let $A(1,2)$, $B(6,3)$, $C(8,7)$ and $D(3,6)$. Prove that $ABCD$ is a parallelogram.

The midpoint of diagonal $AC$ is

$$\left(\frac{1+8}{2},\frac{2+7}{2}\right) =\left(\frac92,\frac92\right).$$

The midpoint of diagonal $BD$ is

$$\left(\frac{6+3}{2},\frac{3+6}{2}\right) =\left(\frac92,\frac92\right).$$

The diagonals therefore bisect each other. Hence $ABCD$ is a parallelogram.

This argument proves more than a sketch could. It uses a defining property of parallelograms.

Worked example 10: find a point from equal distances

Point $P(k,4)$ is equidistant from $A(-1,1)$ and $B(5,3)$. Find $k$.

Equate squared distances:

$$PA^2=(k+1)^2+(4-1)^2=(k+1)^2+9,$$ $$PB^2=(k-5)^2+(4-3)^2=(k-5)^2+1.$$

Therefore

$$(k+1)^2+9=(k-5)^2+1.$$

Expand and simplify:

$$k^2+2k+1+9=k^2-10k+25+1,$$ $$2k+10=-10k+26,$$ $$12k=16,$$ $$\boxed{k=\frac43}.$$

The $k^2$ terms cancel because the set of points equidistant from two fixed points is a straight line, the perpendicular bisector of $AB$.

A reliable problem solving routine

For a coordinate geometry problem:

1. Label every point clearly.
2. Identify the required object: a point, signed change, midpoint or length.
3. Write the relevant formula before substituting.
4. Keep one subtraction order throughout each calculation.
5. Use brackets around negative coordinates.
6. Keep exact surds unless approximation is requested.
7. Check the result against the diagram’s broad geometry, but never trust a diagram as proof.

For example, writing $5-(-2)$ is safer than trying to manage two signs mentally.

Self-check

Try these without looking back.

1. In which quadrant is $(-6,-1)$? What is its reflection in the $y$ axis?
2. Find the signed horizontal and vertical changes from $A(4,-3)$ to $B(-2,5)$.
3. Find the midpoint of $P(-8,7)$ and $Q(6,-5)$.
4. The midpoint of $A(-3,8)$ and $B$ is $(2,1)$. Find $B$.
5. Find the exact distance between $C(-2,-1)$ and $D(4,7)$.
6. Point $R$ is two fifths of the way from $A(1,-3)$ to $B(11,7)$. Find $R$.
7. Points $X(0,1)$, $Y(4,4)$ and $Z(7,0)$ form a triangle. Show that it is isosceles.
8. Point $(t,0)$ is equidistant from $(-2,3)$ and $(4,1)$. Find $t$.

Answers

1. Quadrant III. Reflection in the $y$ axis: $(6,-1)$.
2. $\Delta x=-2-4=-6$ and $\Delta y=5-(-3)=8$.
3. $\left(\dfrac{-8+6}{2},\dfrac{7-5}{2}\right)=(-1,1)$.
4. $B=(2(2)-(-3),2(1)-8)=(7,-6)$.
5. $CD=\sqrt{6^2+8^2}=10$.
6. $B-A=(10,10)$, so $R=(1,-3)+\dfrac25(10,10)=(5,1)$.
7. $XY^2=4^2+3^2=25$ and $YZ^2=3^2+(-4)^2=25$, so $XY=YZ$.
8. $(t+2)^2+9=(t-4)^2+1$, giving $12t=4$ and $t=\dfrac13$.

What to learn next

Continue to straight-line graphs to turn coordinate changes into gradients and equations. Then study circle geometry, where distance from a fixed centre becomes the defining idea.

For the full A-level treatment, including perpendicular lines and intersections, see straight lines in coordinate geometry. Coordinate displacement also becomes more powerful in vectors.