If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Geometry (FL B.E.S.T.)>Unit 1

Lesson 3: Distance and midpoints

# Distance formula

Learn how to find the distance between two points by using the distance formula, which is an application of the Pythagorean theorem. We can rewrite the Pythagorean theorem as d=√((x_2-x_1)²+(y_2-y_1)²) to find the distance between any two points. Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

• OK, this helps a lot, but what about when the triangle does not have a right angle and it's an isosceles triangle or any other triangle?
• when dealing with graphs, this is automatically a right triangle. trig can (with a little geometry) be applied to acute or obtuse triangles.
• What should you do when you are asked to find the distance between a point and a liner equation?
• To be a bit more detailed:
1) You solve the original line equation for y if it isn't already.
2) The perpendicular line to that will be the most direct route to your point. Just take the negative inverse (if your line has a slope of 2, the negative inverse is -1/2). Which will be the slope of your perpendicular line.
3) To find the y-intercept of the perpendicular line you align it with the point you are given (if you have P(2|3) and a slope of -1/2 you can solve y=mx+c for c: 3=-1/2*2+c => c=4 and the perpendicular line will be y=-1/2x+4)
4) Then setting both lines equal you can find out where they intersect, which gets you the second point.
5) Finally you can find out the distance with Pythagoras with the distance between the points as the hypotenuse.
That's the mechanics. If you understand why you do that you have figured out almost all about linear equations.
• What is delta? (To be a little more specific.)
• Delta is a greek letter that in this case stands for change.
Delta x is the change in x. If the first point (3, 1) and the second point is (1,1), then delta x is the change in x is 3-1 or 2. Delta y is the change in y is 1=1 or 0.
• Can someone please tell me what hypotenuse means?
• In geometry, a hypotenuse is the longest side of a right-angled triangle, which is the side opposite the right angle. The length of the hypotenuse of a right triangle can be found using the Pythagorean theorem, which states that the square of the length of the hypotenuse equals the sum of the squares of the lengths of the other two sides. For example, if one of the other sides has a length of 3 metres (when squared, 9 m²) and the other has a length of 4 m (when squared, 16 m²), then their square
• Can you also use rise-over-run for this?
• Rise over run is the formula that basically describes the slope. Slope can also be calculated using the y2 - y1 / x2 - x1 formula. However, the distance formula is different. It's used to describe the length of a line segment or the distance between 2 points. Since the two formulas are used for 2 completely different things, you wouldn't be able to replace one with the other.

Good question, though. Hope this made sense!
• You do not explain how the distance formula, (d=√((x_2-x_1)²+(y_2-y_1)²), relates to the Pythagorean Theorem. Sure, I can understand that it has a relation to the Pythagorean. Nonetheless, it doesn't explain to me how they correspond, espically with the, y Sub,2 items within the formula on its own. That doesn't make any coherent sense to me. No one in my entire time learning Mathematics, has ever explained to me how it relates to the Pythagorean Theorem. Maybe because there is no realistic way to explain it. Maybe I'm just confused or just being way too inquisitive. I don't know. Sal, please explain it to me. I do not know how this question will look when it gets posted, or if you will even happen to notice it. Someone else may answer the inquiry, it may not even be you. I would like you to answer. But regardless, the person does not matter to me. I want to understand how the Distance Formula relates to the Pythagorean Theorem. Although, the formula may be answering itself. I don't know. I do not want to draw random conclusions. If this question is way too long to read, I apologize, I just want to make sure I can get this answered. I know this will help me when I join the Navy later in life. I hope that someone can at least answer the question. When I saw the preview for this question, I was like,' Holy Smokes!, That is a long question!' I hope it can get answered,it really is a goal I have within the question here.

I hope it can get answered,
Harry, From OoTP
• This is a great question! It’s much better to make logical connections between facts in math, instead of just rote-memorizing them.

Imagine a right triangle with vertices at (x_1, y_1), (x_2, y_1), and (x_2, y_2).

The right angle at is (x_2, y_1), the horizontal leg has length |x_2-x_1|, the vertical leg has length |y_2-y_1|, and the hypotenuse is the distance d between (x_1, y_1) and (x_2, y_2). Since the square of the absolute value of a quantity is the same as square of that quantity, the Pythagorean Theorem leads to the distance formula:

d^2 = |x_2-x_1|^2 + |y_2-y_1|^2 from the Pythagorean Theorem

d^2 = (x_2-x_1)^2 + (y_2-y_1)^2

d = sqrt[(x_2-x_1)^2 + (y_2-y_1)^2].
• What about when one of the points is a variable?
• You can still use the formula, and simplify like you would any other algebraic expression.
• What is the delta?
• Delta is a symbol that means "change" in some quantity.
• At Sal used a calculator to find the root, but in my school, calculator is not allowed. So how do you find a root of a non-perfect square number? let's use the example √133
• If a calculator is not allowed, then your school might allow you to write the exact value directly as it is √133.

For some other roots like √8, you will need to simplify it as 2√2 first.

It'll be quite absurd if your school requires you to find an approximation of a root without letting you to use a calculator.