Lesson Explainer: Length and Area in the Coordinate Plane Mathematics

In this explainer, we will learn how to find the distance between two points on the coordinate plane and find the area between three points that form a triangle.

The concept of finding the distance between points on the coordinate plane is strongly linked to the Pythagorean theorem. We recall that for a triangle with hypotenuse 𝑐 and two shorter sides π‘Ž and 𝑏 the Pythagorean theorem states that π‘Ž+𝑏=𝑐.

This theorem means that if it is possible for us to find the horizontal and vertical distances between two points on the coordinate plane, then finding the overall distance is straightforward.

For instance, suppose we considered the points (3,4) and (βˆ’2,1). If we plot these on the coordinate plane, we can see that we can form a right triangle with a horizontal side, a vertical side, and a hypotenuse, which is the line segment joining the two points. The third vertex is at (3,1) as shown.

We can observe that the horizontal distance is the difference between the π‘₯-coordinates of the two points, that is, 3βˆ’(βˆ’2)=5 units. Similarly, the vertical distance is the difference between the 𝑦-coordinates, which is 4βˆ’1=3 units.

Then, we can find the length of the hypotenuse 𝑐 using the Pythagorean theorem, π‘Ž+𝑏=π‘οŠ¨οŠ¨οŠ¨, where π‘Ž=3 and 𝑏=5: 3+5=𝑐9+25=𝑐34=π‘βˆš34=𝑐.

Therefore, the hypotenuse, which is the distance between the two points, can be written as √34 units.

As we might expect, this is a procedure that can be extended to find the distance between any general points on the coordinate plane. Suppose we instead have points (π‘₯,𝑦) and (π‘₯,𝑦) and want to find the distance between them.

If we follow the procedure above, we can create a right triangle where the hypotenuse is the distance between the two points and the other two sides are parallel to the π‘₯- and 𝑦-axes.

Just as before, we can find the lengths of the smaller sides by calculating the differences between the two π‘₯-coordinates and the two 𝑦-coordinates respectively. The only thing we must be careful of is the fact that these differences could be negative, so we take their absolute values as shown.

We can then apply the Pythagorean theorem to find the hypotenuse, 𝑑, as shown: (π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦)=𝑑(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦)=𝑑.

Note that when using this formula we do not need to use the absolute value symbols because the terms (π‘₯βˆ’π‘₯) and (π‘¦βˆ’π‘¦) are being squared. Any squared number will be positive; hence, we will not have any issues with β€œnegative distances.”

We have now derived a formula to calculate the distance between any two points on the coordinate plane, which we can define formally below. This formula is often referred to as the distance formula.

Formula: The Distance between Two Points on the Coordinate Plane

The distance, 𝑑, between two points with coordinates (π‘₯,𝑦) and (π‘₯,𝑦) is given by 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦).

We will now see how we can apply this formula in the following examples.

Example 1: Using the Coordinates of Two Points to Find the Distance between Them

Find the distance between 𝐿(βˆ’2,βˆ’8) and 𝑀(βˆ’6,βˆ’9), giving your answer in radical form if necessary.


To find the distance between these two points, we recall the distance formula, which allows us to find a distance, 𝑑, between two points, (π‘₯,𝑦) and (π‘₯,𝑦). This distance is given by 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦).

Substituting in (π‘₯,𝑦)=(βˆ’2,βˆ’8) and (π‘₯,𝑦)=(βˆ’6,βˆ’9), we get 𝑑=(βˆ’6βˆ’(βˆ’2))+(βˆ’9βˆ’(βˆ’8))=(βˆ’4)+(βˆ’1)=√16+1=√17.lengthunits

We note that the choice of which coordinates we designate as (π‘₯,𝑦) and (π‘₯,𝑦) does not matter. For example, if we had chosen the coordinates the other way around, the calculation would have been given as 𝑑=(βˆ’2βˆ’(βˆ’6))+(βˆ’8βˆ’(βˆ’9))=√4+1=√16+1=√17.lengthunits

This is due to the fact that each of the differences under the square root is squared, meaning the result will always be positive. This also corresponds to our intuition that a distance can be measured from either end.

Let us now attempt an example where we are given the inverse problem; that is, given the distance between two points, we will need to find an unknown coordinate using the distance formula.

Example 2: Finding an Unknown Coordinate Using the Distance Formula

If 𝐴(βˆ’7,π‘₯) and 𝐡(9,14), where 𝐴𝐡=4√17 length units, find all the possible values of π‘₯.


To find the possible values of π‘₯, we can use the given information about the distance between the two points. We recall that the distance, 𝑑, between two points, (π‘₯,𝑦) and (π‘₯,𝑦), is given by 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦).

Here, we have been given that the value of 𝑑 is 4√17, and we can substitute in the values of the two points (π‘₯,𝑦)=(9,14) and (π‘₯,𝑦)=(βˆ’7,π‘₯). Doing this gives us 4√17=(βˆ’7βˆ’9)+(π‘₯βˆ’14)=(βˆ’16)+(π‘₯βˆ’14)=16+(π‘₯βˆ’14).

Note that we could expand the (π‘₯βˆ’14) term, but it is actually more convenient for it to remain in factored form. Now, we can deal with the square roots by squaring both sides and simplifying, which gives us ο€»4√17=16+(π‘₯βˆ’14)16Γ—17=16+(π‘₯βˆ’14).

We then swap the sides around and take the constant terms to the right-hand side: (π‘₯βˆ’14)=16Γ—17βˆ’16Γ—16=16(17βˆ’16)=16.

We can solve this equation by taking the square root of both sides once more. This time, we should be careful to account for both the positive and the negative roots by using the Β± sign. We have (π‘₯βˆ’14)=±√16π‘₯βˆ’14=Β±4.

This gives us two possible values for π‘₯: π‘₯=10 and π‘₯=18. This means that there are two possible positions for 𝐴, marked by 𝐴(βˆ’7,10) and 𝐴(βˆ’7,18), which both lie on the line π‘₯=βˆ’7, as shown in the diagram below.

So far, we have only investigated examples with two points, but there are many more applications of the distance formula that we can use. For instance, if we have the coordinates of the vertices of a shape (such as a triangle), then we can find the lengths of its sides by using the distance formula.

Suppose we were given a triangle with vertices 𝐴(1,2), 𝐡(4,6), and 𝐢(6,1.5), as shown below.

If we were asked whether this is an equilateral triangle or not, would it be possible for us to find this out with the information given to us? Recall that an equilateral triangle is a triangle where all its sides are the same length. At first, it appears as though this is the case, but this is something we ought to verify algebraically to make sure. Specifically, if we use the distance formula on each pair of vertices, we can find the length of each side and compare the lengths to find whether they are truly equal.

We will not demonstrate the full calculation for each pair of points this time, but if we were to find these distances and mark them on the diagram, we would find the following.

That is to say, all of the sides are approximately equal to 5, but they are not the same. Therefore, the triangle is not equilateral (in fact, it is scalene, since none of the sides are the same length and no angle is a right angle).

Let us work through a full example of classifying a triangle by finding the lengths of its sides.

Example 3: Classifying a Triangle Using the Lengths of Its Sides

A triangle has vertices at the points 𝐴(4,2), 𝐡(6,2), and 𝐢(5,βˆ’1).

  1. Work out the lengths of the sides of the triangle. Give your answers as surds in their simplest form.
  2. What type of triangle is 𝐴𝐡𝐢?


Part 1

To begin with, we need to calculate the side lengths of the triangle. We can do this by finding the distance between each pair of vertices using the distance formula. We recall that the distance, 𝑑, between two points is given by 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦), where (π‘₯,𝑦) and (π‘₯,𝑦) are the coordinates of the points (in either order).

Let us use this formula for each pair of points. For 𝐴(4,2) and 𝐡(6,2), we have 𝐴𝐡=(6βˆ’4)+(2βˆ’2)=√2+0=2.lengthunits

Next, for 𝐡(6,2) and 𝐢(5,βˆ’1), we have 𝐡𝐢=(6βˆ’5)+(2βˆ’(βˆ’1))=√1+3=√10.lengthunits

Finally, for 𝐴(4,2) and 𝐢(5,βˆ’1), we have 𝐴𝐢=(5βˆ’4)+(βˆ’1βˆ’2)=1+(βˆ’3)=√10.lengthunits

Part 2

As we can see, 𝐴𝐡=2, 𝐡𝐢=√10, and 𝐴𝐢=√10 (all in length units). We recall that a triangle with two equal sides and one side of a different length is an isosceles triangle. While not required, we can plot the triangle on the coordinate plane to visualize this.

In conclusion, the side lengths are 𝐴𝐡=2, 𝐡𝐢=√10, and 𝐴𝐢=√10 and the triangle is isosceles.

When it comes to shapes in the coordinate plane, we are also interested in their areas. To find the area of a triangle, 𝐴, recall that we have the formula 𝐴=12π‘β„Ž, where 𝑏 is the base length and β„Ž is the height. If a question asks us to find the area of a triangle, we can once again use the distance formula to help us find the quantities we require. The only complication we need to be careful of is the orientation of the triangle, which can change what values we have to consider for the base and the height of the triangle.

Let us investigate how this is done in the next example.

Example 4: Finding the Area of a Triangle Described Using Lines

The straight line π‘™οŠ§ has the equation 𝑦=π‘₯+6, and the straight line π‘™οŠ¨ has the equation π‘₯+𝑦=30. The lines π‘™οŠ§ and π‘™οŠ¨ intersect at the point 𝐴, and the line π‘™οŠ¨ intersects the π‘₯-axis at the point 𝐡. Calculate the area of triangle 𝐴𝑂𝐡, where 𝑂 is the origin.


Here, we have been asked to find the area of a triangle, although the only coordinate that has been given to us directly is the vertex 𝑂, which denotes the origin (0,0). Let us begin by finding the other two vertices of the triangle, 𝐴 and 𝐡.

To find 𝐴, we need to find the intersection point of the lines π‘™οŠ§ and π‘™οŠ¨, which we can do by solving them simultaneously. One way to do this is by substitution. Since we have 𝑦=π‘₯+6, we can substitute this value of 𝑦 directly into π‘₯+𝑦=30 to get π‘₯+(π‘₯+6)=302π‘₯+6=302π‘₯=24π‘₯=12.

Then, we can find 𝑦 by substituting π‘₯=12 into either equation. Taking the first equation, 𝑦=π‘₯+6, we have 𝑦=12+6𝑦=18.

So, 𝐴=(12,18), which we can verify is correct by substitution into the second equation.

Now, let us find 𝐡, which is the point where π‘™οŠ¨ intercepts the π‘₯-axis. We can calculate this by substituting 𝑦=0 into the equation π‘₯+𝑦=30, which immediately gives us π‘₯=30. So, 𝐡=(30,0).

Now that we have the three points, let us sketch them on the coordinate plane so we can understand how the triangle is oriented.

While the lines π‘™οŠ§ and π‘™οŠ¨ have also been drawn for additional clarity, we note that they do not necessarily have to coincide with the sides of the triangle. For the purposes of this question, we only need to consider the locations of the vertices of the triangle.

Now, we need to find the area of the triangle, which we can recall is given by the formula area=12π‘β„Ž, where 𝑏 is the base length and β„Ž is the height of the triangle. While it is possible for us to consider any of the three sides as the base of the triangle, it makes the most sense to take the base along the π‘₯-axis in this instance, since the height will be the vertical distance of 𝐴 from the π‘₯-axis. We highlight this below.

So, the height is just the 𝑦-coordinate of 𝐴(12,18), which is 18, and the base length is the length 𝑂𝐡. In fact, since 𝑂 lies on the 𝑦-axis, we can say that 𝑂𝐡 is equal to the π‘₯-coordinate of 𝐡(30,0), which is 30. Substituting these values into the area formula, we get areasquareunits=12Γ—18Γ—30=270.

In the previous example, we had the advantage of working with a triangle where the base and height were easy to find, to the point where we did not actually need the distance formula. Let us consider a slightly more difficult case where we will have to calculate these values explicitly using the formula.

Example 5: Finding the Area of a Triangle Described Using Lines

Two points 𝑅(4,βˆ’2) and 𝑆(7,0) lie on the line π‘™οŠ§ as shown in the diagram.

The line π‘™οŠ¨ is perpendicular to π‘™οŠ§, passes through point 𝑅, and intersects the 𝑦-axis at the point 𝑇. Calculate the area of triangle 𝑅𝑆𝑇 approximated to two decimal places.


In this example, we have been given some information about a triangle and its vertices and need to find its area. Recall that the area of a triangle 𝐴 is given by 𝐴=12π‘β„Ž, where 𝑏 is the base length and β„Ž is the height. Since we have been given a diagram of the problem, we can begin by annotating the triangle and the quantities we need to calculate.

Thus, we can see that the base, 𝑏, is 𝑅𝑆, and the height, β„Ž, is 𝑅𝑇 (note that there are other sides we could choose to be the base, but this is probably the most straightforward choice). Since we have been given 𝑅(4,βˆ’2) and 𝑆(7,0), we can find the distance 𝑑 between them using the distance formula, 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦), where (π‘₯,𝑦) and (π‘₯,𝑦) are the points in question. Substituting (π‘₯,𝑦)=(4,βˆ’2) and (π‘₯,𝑦)=(7,0), we have 𝑅𝑆=(7βˆ’4)+(0βˆ’(βˆ’2))=√3+2=√13.lengthunits

Now, let us find 𝑅𝑇. We have not explicitly been given the point 𝑇, so we will need to calculate it using the information given to us. We have been told that it is on a line perpendicular to π‘™οŠ§, so we should begin by finding the slope of π‘™οŠ§, which will allow us to find the equation of π‘™οŠ¨.

Since we have two points 𝑅(4,βˆ’2) and 𝑆(7,0) on π‘™οŠ§, its slope π‘š is given by π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯=0βˆ’(βˆ’2)7βˆ’4=23.

Next, we recall that a perpendicular line has slope of βˆ’1π‘š. Therefore, π‘™οŠ¨ has a slope of βˆ’32.

Then, we can find the equation of π‘™οŠ¨ using the point–slope form of a line, π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯), by taking the point (π‘₯,𝑦)=𝑅(4,βˆ’2) and the slope π‘š=βˆ’32: π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯)π‘¦βˆ’(βˆ’2)=βˆ’32(π‘₯βˆ’4)𝑦+2=βˆ’32π‘₯+6𝑦=βˆ’32π‘₯+4.

Now, we can find the point 𝑇 by determining where this line intercepts the 𝑦-axis. This is when π‘₯=0, giving us 𝑦=4.

Finally, we can find the length 𝑅𝑇. Taking (π‘₯,𝑦)=(4,βˆ’2) and (π‘₯,𝑦)=(0,4), we have 𝑅𝑇=(0βˆ’4)+(4βˆ’(βˆ’2))=(βˆ’4)+6=√52=2√13.lengthunits

We now return to the formula for the area of a triangle. Substituting β„Ž=2√13 and 𝑏=√13, we have 𝐴=12√13Γ—2√13=12Γ—2Γ—ο€»βˆš13=13.squareunits

Let us finish by discussing the key points we have learned in this explainer.

Key Points

  • The distance, 𝑑, between two points with coordinates (π‘₯,𝑦) and (π‘₯,𝑦) is given by 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦).
  • If we are given the distance between two points and one of the points has an unknown coordinate, then there can be up to two possible values for this coordinate.
  • We can find the side lengths of shapes by using the distance formula.
  • We can find the area of a triangle 𝐴, using the formula 𝐴=12π‘β„Ž, where 𝑏 is the base length and β„Ž is the height, by calculating the necessary lengths using the distance formula.

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