Distance and Midpoint | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Distance is Pythagoras

Join $P(x_1,y_1)$ to $Q(x_2,y_2)$ . The horizontal change is $x_2-x_1$ and the vertical change is $y_2-y_1$ . Their sizes are the short sides of a right-angled triangle. $PQ$ is the hypotenuse: the side opposite the right angle. Pythagoras adds the squares of the two short sides.

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

A coordinate change can be negative; a length cannot. Squaring removes the sign. Keep the same subtraction order for both coordinates.

Swipe left or right to see the whole diagram →

Distance uses the two coordinate gaps as the short sides of a right triangle. The midpoint averages the two endpoints.

Midpoint means halfway

The has an $x$ -coordinate halfway between the two endpoint $x$ -coordinates, and the same is true for $y$ . Halfway between two numbers is their average.

$M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

Worked example

Find the distance and midpoint between P(-2, 3) and Q(4, -5)

The coordinate gaps are $4-(-2)=6$ and $-5-3=-8$ .
Use Pythagoras: $PQ=\sqrt{6^2+(-8)^2}=\sqrt{100}=10$ .
Average each coordinate: $M=\left(\frac{-2+4}{2},\frac{3-5}{2}\right)=(1,-1)$ .

Answer

Distance: 10

Midpoint: (1, -1)

When a coordinate is unknown

Given the midpoint: equate the x-coordinates, then the y-coordinates. For $P(-7,4)$ , $Q(a,b)$ and midpoint $M(\frac32,-11)$ (Coursebook Worked example 3.1):

$\frac{-7+a}{2}=\frac32\quad\Rightarrow\quad a=10$

$\frac{4+b}{2}=-11\quad\Rightarrow\quad b=-26$

Given the distance: square the coordinate gaps. Preserve both quadratic roots unless the geometry rejects one.

Worked example

The distance from $P(-2,a)$ to $Q(a-2,-7)$ is $17$ . Find $a$

Coursebook Worked example 3.3. Squaring the distance equation removes the square root without changing the positive length.

$(a-2-(-2))^2+(-7-a)^2=17^2$
Expand and simplify: $2a^2+14a-240=0$ , then divide by 2: $a^2+7a-120=0$ .
An ordinary calculator polynomial solver gives the convenient roots $8$ and $-15$ . Show the supported working as $(a-8)(a+15)=0$ .

Answer

a=8\quad\text{or}\quad a=-15

Common mistake

Do not measure a length from a sketch. A coordinate-geometry answer needs calculation; a diagram may not be drawn to scale.

Check the method

Your turnIndependent check

The points are $A(-3,4)$ , $B(5,-2)$ and $C(1,6)$ . Find the midpoint of $AB$ . Then decide whether $CA=CB$ by comparing squared distances.

Show worked answer

Answer

$M=\left(\frac{-3+5}{2},\frac{4-2}{2}\right)=(1,1)$

$CA^2=(-3-1)^2+(4-6)^2=20$ , while $CB^2=(5-1)^2+(-2-6)^2=80$ . Therefore $CA\ne CB$ .

Distance is Pythagoras

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

A coordinate change can be negative; a length cannot. Squaring removes the sign. Keep the same subtraction order for both coordinates.

Swipe left or right to see the whole diagram →

Distance uses the two coordinate gaps as the short sides of a right triangle. The midpoint averages the two endpoints.

Midpoint means halfway

The has an $x$ -coordinate halfway between the two endpoint $x$ -coordinates, and the same is true for $y$ . Halfway between two numbers is their average.

$M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

Worked example

Find the distance and midpoint between P(-2, 3) and Q(4, -5)

The coordinate gaps are $4-(-2)=6$ and $-5-3=-8$ .
Use Pythagoras: $PQ=\sqrt{6^2+(-8)^2}=\sqrt{100}=10$ .
Average each coordinate: $M=\left(\frac{-2+4}{2},\frac{3-5}{2}\right)=(1,-1)$ .

Answer

Distance: 10

Midpoint: (1, -1)

When a coordinate is unknown

Given the midpoint: equate the x-coordinates, then the y-coordinates. For $P(-7,4)$ , $Q(a,b)$ and midpoint $M(\frac32,-11)$ (Coursebook Worked example 3.1):

$\frac{-7+a}{2}=\frac32\quad\Rightarrow\quad a=10$

$\frac{4+b}{2}=-11\quad\Rightarrow\quad b=-26$

Given the distance: square the coordinate gaps. Preserve both quadratic roots unless the geometry rejects one.

Worked example

The distance from $P(-2,a)$ to $Q(a-2,-7)$ is $17$ . Find $a$

Coursebook Worked example 3.3. Squaring the distance equation removes the square root without changing the positive length.

$(a-2-(-2))^2+(-7-a)^2=17^2$
Expand and simplify: $2a^2+14a-240=0$ , then divide by 2: $a^2+7a-120=0$ .
An ordinary calculator polynomial solver gives the convenient roots $8$ and $-15$ . Show the supported working as $(a-8)(a+15)=0$ .

Answer

a=8\quad\text{or}\quad a=-15

Common mistake

Do not measure a length from a sketch. A coordinate-geometry answer needs calculation; a diagram may not be drawn to scale.

Check the method

Your turnIndependent check

The points are $A(-3,4)$ , $B(5,-2)$ and $C(1,6)$ . Find the midpoint of $AB$ . Then decide whether $CA=CB$ by comparing squared distances.

Show worked answer

Answer

$M=\left(\frac{-3+5}{2},\frac{4-2}{2}\right)=(1,1)$

$CA^2=(-3-1)^2+(4-6)^2=20$ , while $CB^2=(5-1)^2+(-2-6)^2=80$ . Therefore $CA\ne CB$ .