The distance between points and the midpoint of a segment

Geometry, grade 10 (4 units) Lesson 1 of 20Learning

The distance between points and the midpoint of a segment

A segment parallel to an axis

Every point in the coordinate system has two coordinates: $\displaystyle x$ – how far right or left, and $\displaystyle y$ – how far up or down. When two points share a coordinate, the segment between them is parallel to one of the axes, and its length is the difference between the other coordinates.

$\displaystyle A(1,3)$ and $\displaystyle B(6,3)$ have the same $\displaystyle y$ , so $\displaystyle AB$ is parallel to the $\displaystyle x$ -axis: $\displaystyle {AB}={6-1}={5}$ .
$\displaystyle B(6,3)$ and $\displaystyle C(6,-1)$ have the same $\displaystyle x$ , so $\displaystyle BC$ is parallel to the $\displaystyle y$ -axis: $\displaystyle {BC}={3-(-1)}={4}$ .
Take the smaller from the larger: a length is never negative.

The distance between two points

A segment that is not parallel to the axes is the hypotenuse of a right-angled triangle whose legs are parallel to the axes. So Pythagoras’ theorem gives its length.

In the drawing, $\displaystyle A(1,1)$ and $\displaystyle B(5,4)$ , and the point $\displaystyle C(5,1)$ completes a right-angled triangle:

the leg $\displaystyle AC$ is parallel to the $\displaystyle x$ -axis: $\displaystyle {AC}={5-1}={4}$ – the difference between the $\displaystyle x$ -coordinates;
the leg $\displaystyle BC$ is parallel to the $\displaystyle y$ -axis: $\displaystyle {BC}={4-1}={3}$ – the difference between the $\displaystyle y$ -coordinates;
by Pythagoras’ theorem: $\displaystyle {AB}={\sqrt{4^{2}+3^{2}}}={\sqrt{25}}={5}$ .

So for any two points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$ , the legs are the differences between the coordinates, and the distance is

\displaystyle d

\displaystyle =

\displaystyle \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

The order of the points makes no difference: each difference is squared, and the square of a negative number is positive.

Example: a distance that is a root

The distance between $\displaystyle P(-2,3)$ and $\displaystyle Q(4,-1)$ :

PQ=\sqrt{(4-(-2))^{2}+(-1-3)^{2}}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}

$\displaystyle {\sqrt{52}}={\sqrt{4\cdot13}}={2\sqrt{13}}$ : the $\displaystyle 4$ comes out from under the root.
Leave the result as a root. $\displaystyle 7.21$ is only an approximation of it, and working on with it gives results that are not exact.

Exercise 1

Find the distance between the two points.
Write the result exactly, in its simplest form.

\displaystyle A(2,1)

\displaystyle B(5,5)

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle C(-1,4)

\displaystyle D(3,-2)

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle E(0,-3)

\displaystyle F(-3,0)

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The midpoint of a segment

The coordinates of a segment’s midpoint are the means of the coordinates of its ends: its $\displaystyle x$ is halfway between the two $\displaystyle x$ -coordinates, and its $\displaystyle y$ halfway between the two $\displaystyle y$ -coordinates.

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

In the drawing, the midpoint of the segment from $\displaystyle A(2,-3)$ to $\displaystyle B(8,5)$ :

$\displaystyle {x_M}={\frac{2+8}{2}}={5}$ ;
$\displaystyle {y_M}={\frac{-3+5}{2}}={1}$ ;
so $\displaystyle M(5,1)$ . A check: from $\displaystyle A$ to $\displaystyle M$ is $\displaystyle 3$ to the right and $\displaystyle 4$ up, and so is $\displaystyle M$ to $\displaystyle B$ .

Exercise 2

Find the midpoint of the segment.

\displaystyle A(-3,5)

\displaystyle B(7,1)

\displaystyle x_M

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle y_M

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle C(-4,-6)

\displaystyle D(1,0)

\displaystyle x_M

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle y_M

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

One end of a segment from its midpoint

When one end and the midpoint of a segment are known, the other end comes from an equation: the midpoint formula, with the missing coordinate as its unknown.

Example: $\displaystyle M(3,1)$ is the midpoint of the segment $\displaystyle AB$ , and $\displaystyle A(-1,4)$ .

$\displaystyle {\frac{-1+x}{2}}={3}$ , so $\displaystyle {-1+x}={6}$ and $\displaystyle {x}={7}$ ;
$\displaystyle {\frac{4+y}{2}}={1}$ , so $\displaystyle {4+y}={2}$ and $\displaystyle {y}={-2}$ ;
$\displaystyle B(7,-2)$ .

Exercise 3

The point $\displaystyle M(2,-1)$ is the midpoint of the segment $\displaystyle PQ$ .
$\displaystyle P(5,3)$ .
Find the coordinates of the point $\displaystyle Q$ .

\displaystyle x_Q

\displaystyle y_Q

Exercise 4

The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(1,1)$ , $\displaystyle B(5,3)$ and $\displaystyle C(3,7)$ .

Find the lengths of the sides.

\displaystyle AB

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle BC

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle AC

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

What kind of triangle is $\displaystyle ABC$ ?
Explain by the lengths you found.

Exercise 5

$\displaystyle ABCD$ is a parallelogram.
Three of its vertices are $\displaystyle A(-4,-3)$ , $\displaystyle B(-1,2)$ and $\displaystyle C(5,4)$ .
A parallelogram’s diagonals bisect each other, so the diagonals $\displaystyle AC$ and $\displaystyle BD$ have the same midpoint.

Find the coordinates of the vertex $\displaystyle D$ .

\displaystyle x_D

\displaystyle y_D

Find the lengths of the sides $\displaystyle AB$ and $\displaystyle BC$ .

\displaystyle AB

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle BC

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: geometry, grade 10 (4 units)

Open as a page

Analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
A line’s equation in explicit form – $\displaystyle m$ the slope, $\displaystyle b$ where it meets the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
A line’s equation in implicit form
$\displaystyle Ax+By+C=0$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
The slope – $\displaystyle \alpha$ the angle between the line and the positive direction of the $\displaystyle x$ -axis
$\displaystyle m$ $\displaystyle =$ $\displaystyle \tan\alpha$

Similar triangles and midsegments

The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
The ratio of the perimeters
$\displaystyle \frac{P_1}{P_2}$ $\displaystyle =$ $\displaystyle k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$
The ratio of corresponding altitudes, medians and angle bisectors
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{d_1}{d_2}=k$
A triangle’s midsegment – $\displaystyle DE$ parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$

Trigonometry

Sine – $\displaystyle a$ the side opposite the angle, $\displaystyle c$ the hypotenuse
$\displaystyle \sin\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{c}$
Cosine – $\displaystyle b$ the side adjacent to the angle
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{b}{c}$
Tangent – the opposite side over the adjacent side
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{b}$
The tangent from the sine and the cosine
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
The identity from Pythagoras
$\displaystyle \sin^{2}\alpha+\cos^{2}\alpha$ $\displaystyle =$ $\displaystyle 1$
Angles whose sum is $\displaystyle 90^{\circ}$
$\displaystyle \sin(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
$\displaystyle \cos(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of an obtuse angle
$\displaystyle \sin(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
A triangle’s area – $\displaystyle \gamma$ the angle between the sides $\displaystyle a$ and $\displaystyle b$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$

Exact values

$\displaystyle \sin30^{\circ}=\cos60^{\circ}=\frac{1}{2}$
$\displaystyle \sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}$
$\displaystyle \sin60^{\circ}=\cos30^{\circ}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan30^{\circ}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{3}}{3}$
$\displaystyle \tan45^{\circ}$ $\displaystyle =$ $\displaystyle 1$
$\displaystyle \tan60^{\circ}$ $\displaystyle =$ $\displaystyle \sqrt{3}$

Special lines in a triangle

Where the medians meet, $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Where the medians meet, in coordinates
$\displaystyle G\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$
The median $\displaystyle AD$ splits the triangle into two triangles of equal area
$\displaystyle S_{ABD}$ $\displaystyle =$ $\displaystyle S_{ACD}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$