Coordinate plane

Coordinate plane is a two – dimensional number line where each point is uniquely specified by a pair of numerical coordinates. It contains two vertical lines called axis.

Furthermore, the horizontal line is called the $x$-axis and the vertical one is the $y$-axis. Their intersection is called the point of origin.

Coordinate is an ordered pair which tells us the location of the point it represents. The first number in the ordered pair tells us how far the point is located from the point of origin on the $x$-axis. Similarly, the other number tells us how far the point is located from the point of origin on the $y$-axis.

Coordinates of the point of origin are $(0, 0)$.

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Both axis have a positive and a negative side. Negative side of the $x$-axis is on the left from the point of origin, and positive on the right. Similarly, negative side of the $y$-axis is located below the point of origin, and positive above.

Furthermore, the two axes divide the plane into four sections called quadrants. The quadrants are labelled with Roman numerals, starting at the positive side of $x$-axis and going anti-clockwise.

Points in the coordinate plane

All the points in the first quadrant have both coordinates positive, in the second quadrant all the $x$ coordinates are negative and $y$ coordinates positive, in the third both coordinates are negative, and in the fourth the $x$ coordinates are positive and $y$ coordinates negative.

For example, $A(1,1)$ is a point in the first quadrant, $B(-3,2)$ in the second quadrant, $C(-4,-4)$ in the third quadrant and $D(5,-2)$ in the fourth.

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How to determine coordinates of specific points located in a coordinate plane?

$x$ coordinate is found by drawing a line through $A$ perpendicular to $x$-axis, while $y$ coordinate is found by drawing a line through $A$ perpendicular to $y$-axis.

Furthermore, the vertical line intersects the $x$-axis in $2$, and horizontal line $y$-axis in $3$. As a result, the $x$-coordinate of $A$ is $2$, and $y$-coordinate of $A$ is $3$. Subsequently, this is written as: $A(2, 3)$.

The midpoint formula

How can we calculate the exact middle of the segment $AB$ if, for example, $ A(2, 4)$ and $ B (-4, 2)$?

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Midpoint of a segment is halfway between the two endpoints of the segment. That is to say, its $x$ coordinate is halfway between the $x$ coordinates of the endpoints, while its $y$ coordinate is halfway between the $y$ coordinates of the endpoints.

Therefore, if we want to calculate the midpoint, we simply add modules of $x$ coordinates and divide by 2. After that, we do the same for $y$ coordinates.

Subsequently, in our example: $$\frac{-4 + 2}{2} = -1, \frac{4+2}{2} = 3.$$

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More precisely, the exact formula is:

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

where $x_1$ and $y_1$ are coordinates of the first point, and $x_2$ and $y_2$ of the second.

The distance formula

The distance between points $A$ and $B$ is marked with a modulus: $|AB|$. Furthermore, it represents the shortest length from one point to another.

For instance, if $A(-2, 2), B( 4, -2)$ and $C(4, 2)$, then the distance between $A$ and $C$ is easy to determine since their $y$ coordinates are the same. Therefore, all you have to do is to add length from the left to the one on the right. In other words, $|AC|= 2 + 4 = 6$.

Similarly, $C$ and $B$ share the $x$ coordinate so: $|CB|= 2 + 2 = 4$.

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But, what to do with the distance between $A$ and $B$? As you may have already noticed, triangle $ABC$ is a right – angled triangle. Also, the distance between $A$ and $B$ is the same as the length of its hypotenuse. Therefore, we now know that we can easily calculate the wanted distance from the Pythagorean theorem.

In other words,

$$|AB|^2 = |AC|^2 + |CB|^2$$

$$|AB|^2 = 36 + 16 = 52$$

$$|AB|^2 = \sqrt{52}.$$

However, this is a longer way, so let discuss a shorter one. In other words, if we have points $\ P_1(x_1, y_1)$ and $\ P_2(x_2, y_1)$, then:

$\vert P_1 P_2 \vert = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}.$

(It doesn’t matter which point is used as $P_1$ and which as $P_2$.)

Example 1: Calculate the distance between $A(2, -2)$ and $B(4,-2)$.

Solution:

$\vert AB \vert= \sqrt{(4 – 2)^2 + (-2 – (-2))^2} = \sqrt{2^2 + (-2 + 2)^2} = \sqrt{4 + 0^2} = \sqrt{4} = 2$

Parallel lines in the coordinate plane

A slope of a line is a number that describes steepness of the line, or how much it angles from $x$-axis. We usually denote it by $m$.

In addition, if the slopes of two lines are equal, then the lines are parallel.

Furthermore, we can calculate the slope by using a formula:

$m = \frac{y_2 – y_1}{x_2 – x_1},$

where $T_{1} = (x_1, y_1)$ and $T_{2} = (x_2, y_2)$ are arbitrary points of a given line.

Example 2:

If $A\left (-4, 5 \right)$ and $B\left(\frac{2}{3}, 4 \right)$, calculate the slope of a line $AB$.

Solution:

$m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{4 – 5}{\frac{2}{3} – (-4)} = \frac{-1}{\frac{14}{3}} = – \frac{3}{14}$

In conclusion, the slope of this line equals to $ -\frac{3}{14}$.

Example 3:

Calculate the slope of a given line.

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Solution:

You simply choose any two points which the line goes through. For instance, we can choose $(-2, 0)$ and $(0, 1)$. As a result, we get

$m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{1 – 0}{0 – (-2)} = \frac{1}{2}.$

Example 4:

If the first line contains points $A(2, 4)$ and $B(3, 2)$, and second line points $C(4,8)$ and $D(6, 4)$, are those lines parallel?

Solution:

$m_1 = \frac{y_2 – y_1}{x_2 – x_1} = \frac{2 – 4}{3 – 2} = \frac{2 – 4}{3 – 2} = -2$

Similarly,

$m_2 = \frac{y_2 – y_1}{x_2 – x_1} = \frac{4 – 8}{6 – 4} = \frac{-4}{2} = -2.$

Therefore, their slopes are equal so we conclude that the lines are parallel.

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