The slope and the equation of a line through two points

Finance and economics, grade 12 (3 units) Lesson 10 of 16Learning

The slope and the equation of a line through two points

The slope of a straight road

On the city map, a straight road passes through the junction $\displaystyle A(1,3)$ and the junction $\displaystyle B(5,11)$ .
One unit on each axis is 100 metres.

\displaystyle y=2x+1

\displaystyle A=(1,3)

\displaystyle B=(5,11)

From $\displaystyle A$ to $\displaystyle B$ the road goes $\displaystyle {5-1}={4}$ units east and $\displaystyle {11-3}={8}$ units north.
For every unit east the road climbs $\displaystyle {\frac{8}{4}}={2}$ units north: that is its slope.

The slope of a line through $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$ is the difference between the $\displaystyle y$ -coordinates divided by the difference between the $\displaystyle x$ -coordinates:

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

A positive slope: the line rises from left to right.
A negative slope: the line falls from left to right.

The equation of a line through two points

A check at the second point:

\displaystyle {2\cdot5+1}={11}

, so the point

\displaystyle B(5,11)

is on the line.

The equation of a line is $\displaystyle {y}={mx+b}$ : you find $\displaystyle m$ from the two points, and $\displaystyle b$ by substituting one of them.

The solution below finds the equation of the road through $\displaystyle A(1,3)$ and $\displaystyle B(5,11)$ :

The slope: $\displaystyle {m}={\frac{11-3}{5-1}}={\frac{8}{4}}={2}$ , so $\displaystyle {y}={2x+b}$ .
We substitute $\displaystyle A(1,3)$ : $\displaystyle {3}={2\cdot1+b}$ , so $\displaystyle {b}={1}$ .
The road’s equation: $\displaystyle {y}={2x+1}$ .

The formula of a line with slope $\displaystyle m$ through $\displaystyle (x_1,y_1)$ gives the same equation:

\displaystyle y-y_1=m(x-x_1)

Here: $\displaystyle {y-3}={2(x-1)}$ , that is $\displaystyle {y}={2x-2+3}={2x+1}$ .

A point on the line

A point is on the line when its coordinates satisfy the line’s equation.
Are the stops $\displaystyle C(4,9)$ and $\displaystyle D(6,12)$ on the road $\displaystyle {y}={2x+1}$ ?

$\displaystyle C$ : $\displaystyle {2\cdot4+1}={9}$ , which is the $\displaystyle y$ -coordinate of $\displaystyle C$ : the stop is on the road.
$\displaystyle D$ : $\displaystyle {2\cdot6+1}={13}$ , not 12: the stop is not on the road, it is one unit south of it.

A price as a line through two points

A price made of a fixed cost and a cost per unit is a line too, as in grade 10.
A removals company charged 700 ₪ for a move of 20 km, and 1,000 ₪ for a move of 50 km.

The points on the line: $\displaystyle (20,700)$ and $\displaystyle (50,1000)$ , where $\displaystyle x$ is the distance in km and $\displaystyle y$ the price in shekels.
The slope – the price per km: $\displaystyle {m}={\frac{1000-700}{50-20}}={\frac{300}{30}}={10}$ ₪.
We substitute $\displaystyle (20,700)$ : $\displaystyle {700}={10\cdot20+b}$ , so $\displaystyle {b}={500}$ ₪ – the fixed cost of a move.
The equation: $\displaystyle {y}={10x+500}$ .

Exercise 1

A bus route runs in a straight line through the stop $\displaystyle A(0,2)$ and the stop $\displaystyle B(4,10)$ .

Work out the slope of the route.

/ – 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

Write the route’s equation on the graph board.

\displaystyle A=(0,2)

\displaystyle B=(4,10)

Is the stop $\displaystyle C(6,14)$ on the route?
Substitute its coordinates and explain.

Exercise 2

A straight road passes through the points $\displaystyle P(-2,7)$ and $\displaystyle Q(4,-5)$ .
Work out the slope, and find $\displaystyle b$ .
Write the road’s equation on the graph board.

\displaystyle P=(-2,7)

\displaystyle Q=(4,-5)

Exercise 3

The price of a taxi ride is made of a fixed cost and a cost for every kilometre.
A ride of 6 km cost 40 ₪, and a ride of 10 km cost 60 ₪.
$\displaystyle x$ is the length of the ride in kilometres, and $\displaystyle y$ the price in shekels.

Write the equation of the price on the graph board.

\displaystyle (6,40)

\displaystyle (10,60)

How much will a ride of 15 km cost?

/ – 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 do the slope and $\displaystyle b$ say about the price of the ride?

Exercise 4

A straight water pipe passes through the points $\displaystyle A$ and $\displaystyle B$ on the graph board.
Write the pipe’s equation on the graph board.

\displaystyle y=\,?

\displaystyle A=(-2,0)

\displaystyle B=(4,3)

Formulas: finance and economics, grade 12 (3 units)

Open as a page

Revenue and profit: a quadratic model

The revenue – $\displaystyle p$ the price of a unit, $\displaystyle q$ the number of units sold
$\displaystyle R=p\cdot q$
The profit – the revenue minus the cost
$\displaystyle P=R-C$
A parabola – the revenue or the profit as a function of the price
$\displaystyle y=ax^{2}+bx+c$
The vertex’s $\displaystyle x$ – where the revenue or the profit is greatest (when $\displaystyle a<0$ )
$\displaystyle x=-\frac{b}{2a}$ $\displaystyle a\neq0$
The quadratic formula – the prices where the revenue or the profit is 0
$\displaystyle x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Linear programming

A constraint – an inequality; the points that satisfy it are a half-plane
$\displaystyle ax+by\le c$
The objective function – the greatest profit is at one of the feasible region’s corner points
$\displaystyle P=ax+by$

Analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
The midpoint of the segment between the two points
$\displaystyle M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
The slope of the line through the two points
$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$
The equation of the line through the point $\displaystyle (x_1,y_1)$ with slope $\displaystyle m$
$\displaystyle y-y_1=m(x-x_1)$
Parallel lines – the same slope
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$

The normal distribution

In a normal distribution – $\displaystyle \mu$ the mean, $\displaystyle \sigma$ the standard deviation: about 68% of the data within one standard deviation of the mean (34% on each side)
$\displaystyle P(\mu-\sigma<x<\mu+\sigma)\approx68\%$
About 95% within two standard deviations of the mean (another 13.5% on each side)
$\displaystyle P(\mu-2\sigma<x<\mu+2\sigma)\approx95\%$
About 2.5% above $\displaystyle \mu+2\sigma$ , and about 2.5% below $\displaystyle \mu-2\sigma$
$\displaystyle P(x>\mu+2\sigma)\approx2.5\%$
The standard score of the value $\displaystyle x$ – how many standard deviations it is above the mean (or below it, when $\displaystyle z<0$ )
$\displaystyle z=\frac{x-\mu}{\sigma}$
The value whose standard score is $\displaystyle z$
$\displaystyle x=\mu+z\cdot\sigma$