The slope and the equation of a line

Geometry, grade 10 (4 units) Lesson 2 of 20Learning

The slope and the equation of a line

The slope from two points

The slope of a line is the ratio of the change in $\displaystyle y$ to the change in $\displaystyle x$ , going from one point of the line to another. The ratio is the same for any two points of the line, so any two points give it.

\displaystyle m

\displaystyle =

\displaystyle \frac{y_2-y_1}{x_2-x_1}

\displaystyle x_1\neq x_2

\displaystyle y=2x-1

\displaystyle A=(1,1)

\displaystyle B=(4,7)

In the drawing, the line passes through $\displaystyle A(1,1)$ and $\displaystyle B(4,7)$ :

from $\displaystyle A$ to $\displaystyle B$ , $\displaystyle y$ rises by $\displaystyle {7-1}={6}$ , and $\displaystyle x$ grows by $\displaystyle {4-1}={3}$ ;
$\displaystyle {m}={\frac{6}{3}}={2}$ : for each step to the right, the line rises $\displaystyle 2$ .

The sign of the slope

The sign of the slope says which way the line goes, read from left to right:

\displaystyle y=x+1

\displaystyle y=-\frac{1}{2}x+3

\displaystyle y=-2

\displaystyle x=3

a positive slope – the line goes up ( $\displaystyle {y}={x+1}$ );
a negative slope – the line goes down ( $\displaystyle {y}={-\frac{1}{2}x+3}$ );
a slope of $\displaystyle 0$ – the line is parallel to the $\displaystyle x$ -axis: all its points have the same $\displaystyle y$ , and its equation is $\displaystyle {y}={-2}$ ;
a line parallel to the $\displaystyle y$ -axis: all its points have the same $\displaystyle x$ , and its equation is $\displaystyle {x}={3}$ . There $\displaystyle x$ does not change, the denominator in the slope formula is $\displaystyle 0$ , so such a line has no slope.

Exercise 1

Find the slope of the line through the two points.

\displaystyle (2,3)

\displaystyle (6,11)

/ – 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 (-1,4)

\displaystyle (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 (0,5)

\displaystyle (4,5)

/ – 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 explicit form and the implicit form

A line’s equation is written in one of two forms:

explicit form – $\displaystyle y$ alone on one side; $\displaystyle m$ is the slope, and $\displaystyle b$ the $\displaystyle y$ -coordinate of the point where the line crosses the $\displaystyle y$ -axis:

\displaystyle y

\displaystyle =

\displaystyle mx+b

implicit form – all the terms on one side:

\displaystyle Ax+By+C

\displaystyle =

\displaystyle 0

To go from the implicit form to the explicit form, get $\displaystyle y$ on its own:

2x+3y-6=0\quad\Rightarrow\quad 3y=-2x+6\quad\Rightarrow\quad y=-\frac{2}{3}x+2

The line’s slope is $\displaystyle -\frac{2}{3}$ , and it crosses the $\displaystyle y$ -axis at $\displaystyle (0,2)$ .

Exercise 2

The board shows the line $\displaystyle {4x-2y+2}={0}$ .
Write its equation in explicit form.

\displaystyle 4x-2y+2=0

A line from its slope and a point

Through a given point there is only one line of each slope: two lines of the same slope are parallel, and through a point outside a line there is only one line parallel to it. So a slope and a point fix one line, and its equation is:

\displaystyle y-y_1

\displaystyle =

\displaystyle m(x-x_1)

Example: the line of slope $\displaystyle 3$ through $\displaystyle (2,-1)$ :

y-(-1)=3(x-2)\quad\Rightarrow\quad y+1=3x-6\quad\Rightarrow\quad y=3x-7

A check: $\displaystyle {3\cdot2-7}={-1}$ , and the point is on the line.

Exercise 3

The line’s slope is $\displaystyle -2$ , and it passes through the point $\displaystyle P(1,4)$ .
Write the line’s equation in explicit form.

\displaystyle P=(1,4)

A line from two points

Through two points there is one and only one line. Its equation comes in two steps: the slope from the two points, then the line from the slope and one of the points.

Example: the line through $\displaystyle A(-1,5)$ and $\displaystyle B(3,-3)$ :

$\displaystyle {m}={\frac{-3-5}{3-(-1)}}={\frac{-8}{4}}={-2}$ ;
through $\displaystyle A$ : $\displaystyle {y-5}={-2(x+1)}$ , so $\displaystyle {y}={-2x+3}$ ;
a check at $\displaystyle B$ : $\displaystyle {-2\cdot3+3}={-3}$ .

Exercise 4

Write the equation of the line through the points $\displaystyle A(-2,-1)$ and $\displaystyle B(2,7)$ .

\displaystyle A=(-2,-1)

\displaystyle B=(2,7)

Exercise 5

Write the equation of the line through the points $\displaystyle A(-3,2)$ and $\displaystyle B(5,-2)$ .

\displaystyle A=(-3,2)

\displaystyle B=(5,-2)

A point on a line

A point is on a line when its coordinates satisfy the line’s equation: substitute its $\displaystyle x$ , and you get its $\displaystyle y$ .

$\displaystyle (2,5)$ is on the line $\displaystyle {y}={3x-1}$ : $\displaystyle {3\cdot2-1}={5}$ ;
$\displaystyle (1,4)$ is not on it: $\displaystyle {3\cdot1-1}={2\neq4}$ .

Exercise 6

The line is $\displaystyle {y}={-2x+5}$ .

Is the point $\displaystyle (3,-1)$ on the line?
And the point $\displaystyle (-1,3)$ ?
Explain.

The point $\displaystyle (a,-7)$ is on the line.
Find $\displaystyle a$ .

Formulas: geometry, grade 10 (4 units)

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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 $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$