Lesson Library> Coordinates & Equations
Coordinates & Equations
A Specific Place in a Specific Space
“Hey Jack! Could you hand me that bottle of water?”
“Which bottle do you want?”
“That bottle!”
Sometimes we have to be more specific about what we are trying to say. We might use names, prepositions, and even serial numbers to specify a specific place or thing. And it is the same in mathematics.
In this lesson, we are going to learn everything about coordinate systems! And by the end of this lesson, you will for sure know how to read and use them!
A coordinate system is made up of two number lines that intersect, creating a map. The horizontal number line is called the $x$-axis, and the numbers on it are increasing to the right. The vertical number line is called the $y$-axis, and the number on it is increasing upwards. The intersection is called the origin, and it is a fixed point where the lines form a right angle with each other.
We can pinpoint an exact location on the coordinate system by using coordinates, which is the indexes used to describe a point. A location on the coordinate system is described by coordinates, the following format: \((x, y) \). The \(x \) stands for the $x$-coordinate of the point, and \(y \) stands for the $y$-coordinate.
We can determine the coordinates of a point by simply reading where on the $x$- and the $y$-axis it is located.
Example 1
In the following coordinate system, the point \(A \) has the coordinates: \((2, 4) \), and \(B \) has the coordinates: \((3, -1) \).
What are the coordinates for the point \(C \) ?
Solution
We read on the $x$-axis where the point is located, which is \(-4 \), and it is also located on \(-2 \) on the $y$-axis. Therefore, the point \(C \) has the coordinates: \((-4, -2) \)
In the same way, the origin has the coordinates \((0, 0) \).
Note! Coordinates on the coordinate system can also be negative.
Now let’s move on to the exercises!
Exercise 1
What coordinates do the following points have?
\(A\): \(\mathbf{(2, 4) }\)
\(B\): \(\mathbf{(0, 3)}\)
\(C\): \(\mathbf{(-1.5, 0)} \)
\(D\): \(\mathbf{(3, -1)}\)
\(E\): \(\mathbf{(-2.5, -2.5)}\)
Exercise 2
Which of the following points has these coordinates:
\(a)\) \((3, 3) \)
\(b)\) \((-3, 1) \)
\(c)\) \((2, -1) \)
\(d)\) \((-4, -3) \)
\(e)\) \((-2, 0) \)
The coordinate system is divided into four general sections, called quadrants. The first quadrant is in the upper right corner where both $x$ and $y$ are positive, and is labeled with a Roman numeral \(I\). The order of the other quadrants then go counterclockwise from there.
Notice that in each quadrant, the $x$- and $y$-coordinates will stay either positive or negative:
$$Quadrant\:I:\:(Positive, Positive)$$
$$Quadrant\:II:\:(Negative, Positive)$$
$$Quadrant\:III:\:(Negative, Negative)$$
$$Quadrant\:IV:\:(Positive, Negative)$$
Exercise 3
What quadrant are the following points in?
a) \((1, 5) \)
b) \((-21, -58) \)
c) \((190.23, -42.99) \)
d) \((a, 3a) \), where \(a \) is a positive integer.
e) \((-b, b) \), where \(b \) is a positive integer.
f) \((-c, c) \), where \(c\) is a negative integer.
g) \((\frac{d}{2}, 2d) \), where \(d\) is a negative integer.
Illustrating Equations on a Coordinate System
Some people are just like me, I’m not an artist, and really don’t know how to draw. But drawing linear equations is something that anyone can learn and do!
To explain how to illustrate different equations on a coordinate system, we have to learn how to rewrite equations in a specific way.
In this example, we are going to use the most common two variables: \(x\) and \(y\).
Example 1
How does the variable \(y\) relate to the variable \(x\) in the following equation?
$$2y - 5 = 6x + 3$$
Solution
We could rewrite this equation with algebra:
$$2y - 5 = 6x + 3$$
$$2y = 6x + 8$$
$$\mathbf{y = 3x + 4}$$
Or if we want it written as a function:
$$\mathbf{y(x) = 3x + 4}$$
By rewriting the equation like this, we could now express the equation as y in terms of x, or that y is a function of x. In other words, \(y\) is always \(4\) greater than \(x\) times \(3\).
Exercise 1
Rewrite the following equations as in terms of $x$.
a) \(4y - 12x = 20\)
b) \(3(x+4) = 3y - 51 \)
c) \(z + 2x = \frac{3x + 5}{3}\)
d) \(6(z - x) = 2z + 11 \)
e) \(2(x + y) = x(2x + 3)\)
a) \(y = 3x + 5\)
b) \(y = x + 21\)
c) \(z = -x + \frac{5}{3}\)
d) \(z = \frac{3}{2}x + \frac{11}{4}\)
e) \(y = x^2 + x\)
Because both \(x\) and \(y\) are variables, there is always a variable that is dependent on what value the other variable has.
And by knowing the value of either \(x\) or \(y\), we could solve the corresponding variable to the equation algebraically by substituting that variable with the known value in the function.
Example 1
What is \(y\) if \(x\) is 100 in \(y(x) = 3x + 4 \)? And what is \(x\) if \(y\) is 91 in \(y(x) = 3x + 4 \)?
Solution
We can calculate \(y\) by substituting \(x\) with \(100\):
$$y(x) = 3x + 4 \text{ and } x = 100$$
$$y(100) = 3 \cdot (100) + 4$$
$$y(100) = 304$$
We could also solve it the other way around. If we know that \(y\) is \(91\) :
$$y(x) = 3x + 4 \text{ and } y = 91$$
$$91 = 3x + 4$$
$$87 = 3x$$
$$x = 29$$
Exercise 2
By using the function \(y = 5x - 2\), what is...
a) … the y-value if \(x = 8\)?
b) … the x-value if \(y = 8\)?
c) … \(y(10)\)?
d) … \(y(0)\)?
e) … \(y(-10)\)?
a) \(y = 38\)
b) \(x = 2\)
c) \(y(10) = 5 \cdot (10) -2 = 48\)
d) \(y(0) = -2\)
e) \(y(-10) = -52\)
With the help of the coordinate system, we can easily illustrate relations between different variables.
If a banana costs \(SEK 5 \) each, then we could first write the equation as the following function:
$$ y(x) = 5x $$
By looking at this function, we could say that the total price is always 5 times larger than the number of bananas. And because we wrote the function as y in terms of $x$: How much you pay depends on the number of bananas you buy, or $y$ is a function of $x$.
We could also draw a line on the coordinate system to show the relation between the price of the bananas and the number of bananas we have bought. Later in this lesson, we are going to learn where and how the line is drawn.
We use the $x$-axis for the number of bananas we want to buy, and the $y$-axis for the total price.
This straight line is called a linear equation. With this, we can easily find the corresponding cost for each number of bananas.
If we want to know how much \(7 \) bananas costs in total, we vertically draw a straight line from the number \(7 \) on the x-axis, until we arrive to the linear equation, and then continue by drawing a horizontal line until we reach the $y$-axis. Then we read off the number line to know the price of \(7 \) bananas.
In this case, the price for \(7 \) bananas were \(SEK 35\).
This works because every point on this linear equation is a solution to the equation \(y = 5x\).
For example, \((0, 0)\), \((3, 15)\) and \((7, 35)\).
We can also calculate the value of the graph where it intersects with the axis, known as intercepts.
If we want to calculate the $y$-coordinate of where the graph intersects with the $y$-axis, we could set \( x = 0\). This is the $y$-intercept.
Likewise, if we want to calculate the $x$-coordinate of where the graph intersects with the $x$-axis, we could set \(y = 0\). This is the $x-intercept.
Exercise 3
At a holiday resort you can rent a car for \(SEK 400\). But for each kilometer you drive, it also costs an additional \(SEK 30\).
How much does the car cost in total if you drive \(x \) kilometers?
Let’s call the total cost for \(y \), then we are going to write \(y \) as a function of \(x \), or that the cost of the rental car is a function of the distance you drive.
If we only rent the car, but do not drive the car at all. Then \(y = 400\), because the base-price to rent the car is \(SEK 400 \).
If we drive exactly \(1 km\) with the car, then \(y = 30 \cdot 1 + 400 = 430\).
If we drive \(2 km \) instead, then the cost becomes \(y = 30 \cdot 2 + 400 = 460\).
Because \(x \) is the distance you drive with the car, we can write a general function of \(y \):
$$y = 30x + 400$$
It costs \(\mathbf{30x + 400}\) SEK if you drive\(x \) kilometers.
We can also draw this as a line to visually illustrate it.
Exercise 4
A water container has \(1000 \) ml of water. But on the bottom of the container there is a small crack on it, which makes \(20 \) ml of water fall out constantly every hour.
How much water is left in the container after \(x \) amount of hours?
Let’s use \(y\) as the amount of water left in the container.
At the beginning, when it has gone by \(0 \) hours, \(y = 1000\).
After \(1 \) hour, \(y = 1000 - 20 \cdot 1\).
After \(2 \) hours, \(y = 1000 - 20 \cdot 2 \).
After \(x \) amount of hours, \(y = 1000 - 20 \cdot x \).
Therefore, there is \(\mathbf{1000 - 20 \cdot x }\) or \(\mathbf{ -20 \cdot x + 1000}\) ml of water after \(x \) amount of hours.
And here is a graph to illustrate the linear function \(y = -20 \cdot x + 1000 \):
Linear Equations, Straight Lines
Straight lines are very predictable. When you see a part of it, you will know which direction it is going. But at the same time, straight lines could be infinitely long, and in that way it is also unpredictable.
What is the function for the following linear equation?
This might be hard to figure out sometimes. But at the end of this chapter, you will know how to write down every kind of linear equation.
Every linear equation can be generalized to the form \(y = kx + m\). (Sometimes it is written as \(y = mx + b\))
The \(k \) is called the coefficient, or the slope of the function. It is a constant and explains how much the line will slope on the graph.
The $k$-value of a function is calculated by dividing the rise over the run of the function.
We choose two different points on the function. \(x_1\) and \(y_1\) are the $x$-value and $y$-value of the first point, while \(x_2\) and \(y_2\) are the $ x$-value and $y$-value of the second point.
$$(x_1, y_1)(x_2, y_2)$$
$$k = (\frac{\Delta y}{\Delta x}) = \frac{y_2 - y_1}{x_2 - x_1}$$
The \(m \) is also a constant, and is called the intercept because it tells us where it is intercepting with the $y$-axis.
Example 1
What is the linear equation of this function? (Write it in the form of \(y= kx + m\))
Solution
$$y = kx + m$$
First, we determine the $m$-value by looking for where the equation is intercepting with the $y$-axis at the point \((0, 4)\). Because the $y$-value at that point is \(4\), \((m = 4)\).
$$y = kx + 4$$
After that, we choose two different point that exist on the function. I prefer to pick points that are exact, and not difficult to interpret. For example: \((0, 4)\) and \((3, -5)\).
And then we calculate the $ k$-value:
$$k = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - 4}{3 - 0} = \frac{-9}{3} = -3$$
$$k = -3$$
Therefore, the function is:
$$\mathbf{y = -3x + 4}$$
Exercise 1
Write the following linear functions in the form of \(y= kx + m\).
a) The blue linear function?
b) The pink linear function?
c) The green linear function?
a) \(y=6 x-2\)
b) \(y=-\frac{2}{3}x+3\)
c) \(y=-\frac{5}{3} x\)
Note! Two different linear equations cannot share two of the same points.
This means that if we have two points on a graph, we can always determine the exact linear equation that goes through those points.
Example 1
What is the linear equation that goes through the points in the graph below?
Solution
In the coordinate system above we can see that those two points are \((3, 5)\) and \((6, 12)\). We could draw a straight line that goes through both of them, and calculate the function of that line similar to the last example.
But we could also do it algebraically:
First, we calculate $k$-value:
$$k = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 5}{6 - 3} = \frac{7}{3}$$
$$k = \frac{7}{3}$$
$$y = \frac{7}{3}x + m$$
In this case, we can’t see the $y$-axis, and therefore we cannot figure out the $m$-value by reading off the graph. But we can figure out \(m\) by substituting the $x$-value and the $y$-value with a point that exists on that function.
We know that it has a point at \((3, 5)\), which means that when \(x = 3\), \(y = 5\):
$$5 = \frac{7}{3} \cdot 3 + m$$
$$5 = 7 + m$$
$$m = -2$$
Therefore, the function is:
$$\mathbf{y = \frac{7}{3}x - 2}$$
Exercise 2
Use the information to determine the linear function in the form of \(y = kx + m\).
a) This linear function goes through the points \((21, 6)\) and \((9, 3)\).
b) This linear function goes through the points \((22, 88)\) and \((66, 88)\).
c) This linear function have an m-value of \(19\), and goes through the point \((-3, 9)\).
d) This linear function intersects both x- and y-axis at value of \(-101\).
e) This linear function goes through the points \((-a, a)\) and \((0, 3a)\), where \(a\) is an unknown constant.
a) \(y = \frac{1}{4}x - \frac{3}{4}\)
b) \(y = 0x + 88\) or \(y = 88\)
c) \(y = \frac{10}{3}x + 19\)
d) \(y = -x - 101\)
e) \(y = 2x + 3a\)
Exercise 3
The points \((-9, 247)\), \((5, -47)\) and \((11, a)\) are on the same linear equation.
What is the value of \(a\)?
The k-value of this linear equation is:
$$k = \frac{-47 - 247}{5 - -9} = \frac{-294}{14} = -21$$
$$y = -21x + m$$
And by inserting one of the known points \((5, -47)\), we figure out the $m$-value:
$$y = -21x + m$$
$$-47 = -21 \cdot (5) + m$$
$$m = 58$$
Therefore, the function must be:
$$y = -21x + 58$$
Because \((11, a)\) is a point on this function, \(a\) is the $y$-value of the point where the $x$-value is 11.
By inserting \(x = 11\) into the function, we can calculate the $y$-value.
$$y(11) = -21 \cdot (11) + 58$$
$$y(11) = a$$
$$a = -21 \cdot (11) + 58$$
$$\mathbf{a = -173}$$
Relations Between Linear Functions
Sometimes we can’t see the exact numbers of the linear function, but we want to know about the function quickly. Here are some ways to figure out information about the linear function by just looking at the linear function.
When the $k$-value of a linear function is positive, the $y$-value gets larger as the $x$-value gets larger.
and when the $k$-value is negative, the $y$-value gets smaller as the $x$-value gets larger.
When \(k = 0\), all the different $x$-values have the same $y$-value. This is why in every linear equation, every $x$-value has its own unique $y$-value, as long as \(k\neq0\).
Exercise 1
Do the following linear functions have a k-value that is: \(k > 0\), \(k < 0\) or \(k=0\)?
a) The green linear function?
b) The blue linear function?
c) The purple linear function?
d) The pink linear function?
e) The red linear function?
a) \(k < 0\)
b) \(k = 0\)
c) \(k > 0\)
d) \(k > 0\)
e) \(k < 0\)
If two different linear functions have the same $k$-value, then it implies that those two linear functions are parallel with each other.
$$k_1 = k_2$$
And if two linear functions are perpendicular to each other, the product of their coefficients/k-values of these lines is always \(-1\).
$$k_1 \cdot k_2 = -1$$
Example 1
Find a perpendicular function to \(y = -\frac{3}{5}x + 24\).
Solution
Let’s write the new function like this:
$$y = k_2x + m$$
And because those two lines are perpendicular to each other, we could write this relation between them:
$$-\frac{3}{5} \cdot k_2 = -1$$
Then we can use algebra to calculate the value of \(k_2\).
$$k_2 = \frac{-1}{-\frac{3}{5}}$$
$$k_2 = \frac{5}{3}$$
Therefore, the answer to this question is:
$$\mathbf{y = \frac{5}{3}x + m}$$
And it does not matter which value \(m\) has, because these two functions will stay perpendicular as long as their $k$-values stay in the same relation.
Exercise 2
If a line is parallel to \(y = \frac{7}{8}x + 12\), it must have the same k-value. For that reason, our function must look something like this: \(y = \frac{7}{8}x + m\).
Because it goes through the point \((24, 22)\), we can put in the coordinates as the values for \(x\) and \(y\).
$$22 = \frac{7}{8} \cdot (24) + m$$
$$22 = \frac{7}{8} \cdot (24) + m$$
$$22 = 21 + m$$
$$m = 1$$
Therefore, the function is: \(\mathbf{y = \frac{7}{8}x + 1}\).
Exercise 3
Find the linear function that is perpendicular to \(y = \frac{7}{8}x + 12\) and goes through the origin.
If a line is perpendicular to \(y = \frac{7}{8}x + 12\), the product of the k-values of these lines must be equal to \(-1\).
$$\frac{7}{8} \cdot k_2 = -1$$
$$k_2 = \frac{-1}{\frac{7}{8}}$$
$$k_2 = -\frac{8}{7}$$
$$y = -\frac{8}{7}x + m$$
And because it goes through the origin \((0, 0)\), the m-value must be equal to \(0\).
Therefore, the function is: \(\mathbf{y = -\frac{8}{7}x}\)
Exercise 4
Do the points \((16, 35)\), \((10, 23)\), \((-8, 32)\) and \((-2, 44)\) form a rectangle, if the points are the four corners of the rectangle?
The definition of a rectangle is that it has four right angles and has four sides, which means that it has to have two pairs of lines that are parallel to each other, while the rest are perpendicular to each other.
The linear function that is connecting \((16, 35)\) with \((10, 23)\) is \(y=2x+3\).
The linear function that is connecting \((10, 23)\) with \((-8, 32)\) is \(y=-\frac{1}{2}x+28\).
The linear function that is connecting \((-8, 32)\) with \((-2, 44)\) is \(y=2x+48\).
The linear function that is connecting \((-2, 44)\) with \((16, 35)\) is \(y=-\frac{1}{2}x+43\).
\(y=2x+3\) and \(y=2x+48\) are parallel to each other, as well as \(y=-\frac{1}{2}x+28\) is to \(y=-\frac{1}{2}x+43\).
And the rest that are not parallel to each other are perpendicular. Becuase \(\frac{1}{2} \cdot 2 = -1\).
(The linear functions with the points \((10, 23)\) and \((-2, 44)\) respectively \((16, 35)\) and \((-8, 32)\) are diagonals to the rectangle. It helps a lot by graphing out everything to make it more clear)
Therefore, the points do form a rectangle. The answer is \(\mathbf{\text{yes}}\).
Graphically Solving Equation Systems
Now, when we know how to read and construct linear equations, we are going to solve equation systems graphically!
How do we solve this equation system?
$$\begin{cases} 3x = y +7 \\ 2y = -x + 7 \end{cases}$$
There are a lot of different ways to solve an equation system, but we can solve it with an approach with a coordinate system.
First, we rewrite all the equations to the format of \(y = kx + m\) so that it becomes easier to draw them.
$$\begin{cases} y = 3x - 7 \\ y = -\frac{1}{2}x + \frac{7}{2} \end{cases}$$
Then we can draw these lines by first finding two points that are on the linear equation, and mark them on the coordinate system.
Let’s start with this function: \(y = 3x -7\)
When \(x = 0\), \(y = 3 \cdot (0) - 7 = -7\).
When \(x = 1\), \(y = 3 \cdot (1) - 7 = -4\).
The function \(y = 3x -7\) has points at \((0, -7)\) and \((1, -4)\).
We could write down more points, but we only need two to draw a straight line that goes through them. And it does not matter which points you pick, just choose the points that you think are easy to pinpoint!
And then we do the same to the other linear equation.
We find the solution to the equation system by reading the point of the intersection between these two functions.
In this case, they intersect at the point \((3, 2)\), which means that the solution to this equation system is \(\mathbf{\begin{cases} x = 3 \\ y = 2 \end{cases}}\).
Exercise 1
Find the solution to the following equation system graphically:
$$\begin{cases} 6x + 3(y+2) = 0\\ \frac{2x}{3} - 2y = 4 \end{cases}$$
$$\mathbf{\begin{cases} x = 0\\ y = -2 \end{cases}}$$
The Formula for Calculating a Distance
How far is it between these two points?
We could use the Pythagorean theorem to calculate the distance between the points. We draw a right-angled triangle and use the distance on the x-axis and y-axis between these two points to calculate the distance between the points.
We could also use the following formula, which uses the same method.
$$d = \sqrt {\left( {x_1 - x_2 } \right)^2 + \left( {y_1 - y_2 } \right)^2 }$$
This is called the distance formula, where \(d\) is the distance between the two points, or “the hypotenuse of the triangle”.
Exercise 1
What is the distance between the points \((3, 5)\) and \((9, 13)\)?
$$d = \sqrt {\left( {9-3} \right)^2 + \left( {13-5} \right)^2 } = \sqrt {36 + 64} = 10$$
$$\mathbf{d = 10}$$
Exercise 2
Ebba pointed out four of her favorite points on the coordinate system, showing below. What is the area of the rectangle that the points are forming?
We calculate the area by multiplying the length and the height of the rectangle.
The length of the rectangle is calculated with the distance formula:
$$d = \sqrt {\left( {-8-0} \right)^2 + \left( {16- -16} \right)^2 } = \sqrt {8^2 + 32^2}$$
$$d = \sqrt {1088}$$
The height:
$$d = \sqrt {\left( {20-16} \right)^2 + \left( {8- -8} \right)^2 } = \sqrt {4^2 + 16^2}$$
$$d = \sqrt {272}$$
And the area is:
$$\sqrt {1088} \cdot \sqrt {272} = \mathbf{544}$$
Problems
Exercise 1
The line \(11x + 7ky = 16k\) goes through the point \((20, -5)\). Find \(k\).
Because the linear function \(11x + 7ky = 16k\) goes through the point \((20, -5)\), we know that \(x = 20\) and \(y = -5\) is a solution. We put those values into the function.
$$11 \cdot (20) + 7k \cdot (-5) = 16k$$
$$220 - 35k = 16k$$
$$51k = 220$$
$$k = \mathbf{\frac{220}{51}}$$
Exercise 2
A triangle is formed in the first quadrant bounded by the coordinate axes and a linear graph \(ax + by = 6\), where \(a,b > 0\). If the area of the triangle is \(6\), find the value of \(ab\).
We know that the linear graph \(ax + by = 6\) is forming a triangle in the first quadrant, which means that we can calculate the area of the triangle by knowing the points that the graph intersects with the coordinate axes.
We can calculate the intercepts by setting \(x = 0\), respectively \(y = 0\).
$$ax + by = 6$$ when $$x = 0$$
$$by = 6$$
$$y = \frac{6}{b}$$
$$ax + by = 6$$ when $$y = 0$$
$$ax = 6$$
$$x = \frac{6}{a}$$
This means that the intercepts are \(0, \frac{6}{b}\), and \(\frac{6}{a}, 0\). Which means that the base of the triangle is \(\frac{6}{a}\), and the height is \(\frac{6}{b}\).
$$Area = \frac{Base \cdot Height}{2} = \frac{\frac{6}{a} \cdot \frac{6}{b}}{2}$$
$$Area = \frac{18}{ab}$$
We also know that the area of the triangle is 6.
$$\frac{18}{ab} = 6$$
$$ \mathbf{ab = 3}$$
Exercise 3
Find the two numbers whose sum is 66 and difference is 36.
Let’s call the first number x, and the second number as y. We can create an equation system based on the two relations between these numbers.
$$\begin{cases} x + y = 66 \\ x - y = 36\end{cases}$$
One solution would be drawing these two as functions on a coordinate system, by rewriting them to the format of \(y = kx + m\).
$$\begin{cases} y = 66 - x \\ y = x - 36\end{cases}$$
The functions intersect at the point \((51, 15)\). The solution to the problem is therefore \(\mathbf{\begin{cases} x = 51 \\ y = 15 \end{cases}}\).
Exercise 4
Leo and Michael are taking a walk on a coordinate system. Leo started walking from the point \((-6, 18)\) and went by the point \((2, 20)\). Michael started walking from the point \((28, -6)\) and went by the point \((24, 6)\). They both walk in a straight line. Which point on the coordinate system will their paths meet at?
We start by writing Leo’s path as a linear equation. We know that he went through the points \((-6, 18)\) and \((2, 20)\).
$$y = kx + m$$
$$k = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20 - 18}{2 - (-6)} = \frac{2}{8}$$
$$k = \frac{1}{4}$$
$$y = \frac{1}{4}x + m$$
Then we calculate m by inserting one of the points. \((2, 20)\)
$$20 = \frac{1}{4} \cdot 2 + m$$
$$20 = \frac{1}{2} + m$$
$$m = 19 + \frac{1}{2} = 19.5$$
$$y = \frac{1}{4}x + 19.5$$
Then we can write Michael’s path as another linear equation. We know that he went through the points \((28, -6)\) and \((24, 6)\).
$$y = kx + m$$
$$k = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-6)}{24 - 28} = \frac{12}{- 4}$$
$$k = -3$$
$$y = -3x + m$$
Then we calculate m by inserting one of the points. \((24, 6)\)
$$6 = -3 \cdot 24 + m$$
$$6 = -72 + m$$
$$m = 78$$
$$y = -3x + 78$$
Now we have both of their paths written as functions. And if we draw these two functions on a coordinate system, we can find their intersection.
$$\begin{cases} y = -3x + 78 \\ y = \frac{1}{4}x + 19.5\end{cases}$$
The functions intersect at the point \(\mathbf{(18, 24)}\).
Exercise 5
Calculate the area bounded by the line \(x + y = 10\) and both the coordinate axes.
We start by rewriting the function to the form of \(y = kx + m\).
$$x + y = 10$$
$$y = - x + 10$$
And by drawing this linear equation on a coordinate system, we can see that it is forming a triangle with the axes in the first quadrant.
We can read off the graph that the intercepts are at \((0, 10)\) and \((10, 0)\). To be sure, we can set in the values as \((x = 0)\), and \(y = 0)\) to confirm these points.
$$x + y = 10$$ when $$x = 0$$
$$y = 10$$
$$x + y = 10$$ when $$y = 0$$
$$x = 10$$
Because the linear equation is forming a right-angled triangle, we can just calculate the area with this formula:
$$Area = \frac{Base \cdot Height}{2} = \frac{10 \cdot 10}{2}$$
$$Area = 50$$
The area bounded by the line is \(\mathbf{50}\)
Exercise 6
Kevin and Chengo went to a café to order something to eat. Kevin ordered 3 cupcakes and 2 milkshakes for 35 SEK, while Chengo ordered 9 cupcakes and only 1 milkshake for 55 SEK. How much does a cupcake cost? And how much does a milkshake cost?
Let’s call the price of a cupcake as $x$, and the price of a milkshake as $y$. We can create an equation system based on Kevin and Chengo’s purchases.
$$\begin{cases} 3x + 2y = 35 \\9x + y = 55\end{cases}$$
We then can rewrite these to the format of linear equations.
$$\begin{cases} y = \frac{35}{2} - \frac{3}{2} x \\ y = 55 - 9x\end{cases}$$
Then we can write these equations on a coordinate system, and find the intersection between these two equations.
Exercise 7
Yifei wants to find the $k$-value of a linear equation that goes through the origin. It also goes through another point that is $1$ unit away from the $x$-axis, and $5$ units away from the $y$-axis. This point is in the second quadrant. Help Yifei find the $k$-value of this linear equation.
We know that the equation goes through the origin, which means that it goes through the point \((0, 0)\).
Because the second point is in the second quadrant, we know that the x-coordinate must be negative, while the y-coordinate must be positive. We also know that it is 1 unit away from the x-axis, which means that the y-coordinate must be 1. It is also 5 units away form the y-axis, which means that the x-coordinate must be -5. In that way, the point must be \((-5, 1)\).
Now we can use these two points to calculate the k-value.
$$k = \frac{1 - 0}{-5 - 0} = -\frac{1}{5}$$
The k-value that Yifei was trying to find is \(\mathbf{k = -\frac{1}{5}}\)
Exercise 8
Find \(y\) if \((3, y)\) lies on the line joining \((0, 3/2)\) and \((9/4, 0)\).
We can start by finding the linear equation that goes through the points \((0, 3/2)\) and \((9/4, 0)\).
$$k = \frac{3/2 - 0}{0 - 9/4} = -\frac{12}{18} = -\frac{2}{3}$$
$$y = -\frac{2}{3}x + m$$
By inserting \((0, 3/2)\):
$$3/2 = -\frac{2}{3} \cdot 0 + m$$
$$m = \frac{3}{2}$$
$$y = -\frac{2}{3}x + \frac{3}{2}$$
To find \(y\), we can insert \((3, y)\) into the function.
$$y = -\frac{2}{3} \cdot 3 + \frac{3}{2}$$
$$y = -\frac{1}{2}$$
The answer is \(\mathbf{y = -\frac{1}{2}}\).