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Start by drawing a right triangle in which the hypotenuse will be the segment that connects the points.
See solution.
We want to explain how we can use the Pythagorean Theorem to find the distance between two points on the coordinate plane. Let's start by plotting two random points, A and B.
Next, we will draw a right triangle. The hypotenuse of this triangle will be the segment that connects points A and B. Notice that we can express the length of each of the legs as a difference in x- and y-coordinates of the points.
Now, let's remember the Pythagorean Theorem. This theorem tells us about the relationship between the legs and the hypotenuse.
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Pythagorean Theorem |
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In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. |
Note that the lengths on the legs in our right triangle are x_2-x_1 and y_2-y_1. Let d be the missing distance between the points — the hypotenuse of the triangle. We now have everything to write the Pythagorean Equation. ( x_2-x_1)^2+( y_2-y_1)^2=d^2 Let's now solve the resulting equation for d.
sqrt(LHS)=sqrt(RHS)
Rearrange equation
We see that the result is the Distance Formula. If we substitute the coordinates of the points and evaluate the expression, we get the distance.
Consider the following two points A and B. We want to find the distance d between them.
To find d, we start by drawing a triangle in which the segment joining A and B is the hypotenuse.
Now we can find the lengths of the legs of the triangle by subtracting the x- and y-coordinates of the points. Once we do it, we have everything to solve the triangle.
Let's take the sum of of the squares of the lengths of the legs, in this case 3^2+3^2. From the Pythagorean Theorem we know that it is equal to the square of the length of the hypotenuse, here d^2. 3^2+3^2 = d^2 Let's now solve the resulting equation for d.
Calculate power
Add terms
sqrt(LHS)=sqrt(RHS)
sqrt(a^2)=± a
Use a calculator
Round to 2 decimal place(s)
Rearrange equation
There are two approximate solutions to the equation, d=- 4.24 and d=4.24. Because a distance cannot be negative, d=4.24 is our solution. As a result, the distance between the points is about 4.24 units.