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## Geometry (Eureka Math/EngageNY)

### Course: Geometry (Eureka Math/EngageNY)>Unit 4

Lesson 4: Topic D: Partitioning and extending segments and parameterization of lines

# Distance between point & line

The distance from a point to a line is the shortest distance between the point and any point on the line. This can be done with a variety of tools like slope-intercept form and the Pythagorean Theorem. Created by Sal Khan.

## Want to join the conversation?

• How do i find a point on the line that is NOT the Y intercept. I can see the answer if I graph it, but how do I glean the point on the intercepted line with math. The problem I was working was distance between the point (-6,-5) and the line y=-3x+7. I have the y intercept as -3, but how do I get the intercept on the other line? • Isn't there a formula for this? Can someone explain to me the formula of:
|Ax+By+C| / sqrt(A^2+B^2)
I found it online but it is very vague and doesn't help me. Thanks! • Could you make a video of how to find the distance between 2 parallel lines? • Can't you just use the Pythagorean theorem instead? • Sal does mention ( I forgot which video) that the distance formula is just an application of the Pythagorean Theorem. When you are finding the distance between 2 points, you are essentially trying to find the hypotenuse of those points. Hypotenuse =distance. The change in y over the change in x are the other 2 sides of the triangle.

I like to draw a right triangle and label the side of the 90 degree angle with the change in y (rise) (ex: 3) and the base of the triangle I write down the change in x (run) (ex:2)

Then I just use the Pythagorean Theorem. 3^2 +2^2=c^2
9+4=13
Take the square of 13, can't be simplified. So the distance is square root 13. I like it better that way, cause it's easier to visualize when I draw the triangle.
• Why does making the equations equal create the x coordinate? I'm kind of lost :/ • Wait then can't you use like a graph to find it? Like the f(x) graph since we are trying to find if x equals something y is what? Or am I confusing the 2 together? • Right. Not only that, in some situations, you may be just given the equation of the line and the coordinate, and you have to find the closest value on the line to the point, then calculate the distance. Besides, there's no way a ruler can possibly measure the precise distances. I mean, how will you know exactly what the distance will be if it's square root of 13 or something?
• Why is the slope of the perpendicular line the negative inverse of the slope of the other line? • Here's what I think --- and it will definitely help to draw a graph to illustrate what I say.
1. So draw an original line through the origin that has very little incline (this is just to emphasise what follows). Also sketch its perpendicular, again going through the origin.
2. The slope of the perpendicular line is the opposite sign of the original line because it is going in the opposite direction from the original line (one goes through quadrants 1 and 3, the other through quadrants 2 and 4).
3. The slope of the original line is "change in y" / "change in x". For our purposes, let's denote this as "y original change" / "x original change". Mark it clearly on your graph.
4. The perpendicular line has been formed by rotating the original line by 90 degrees.
Notice that the length of "y perpendicular change" is now the length of "x original change", and similarly, the length of "x perpendicular change" is now the length of "y original change". So the slope of the perpendicular line in now "x original change" / "y original change".
That's why we need to use an inverse for the perpendicular line's slope.
Hope you find this useful!
• I put the points (0,2) and (-2,-6) into the distance formula and I did not get the square root of 40, instead I got the square root of 68? • Find the distance between the point (−1,5) and the line y=1/2x−7.

for this equation can someone please tell me how to get the coordinates of when the line intercepts with the other line.

The equation of the perpendicular line is y=−2x+3.

and also how i would find the 4,-5 without looking in on the graph whats the equation? for something like this?

We can see from the graph (or by setting the equations equal to one another) that the two lines intersect at the point (4,−5). Thus, the distance we're looking for is the distance between the two red points.
(1 vote) • Robert,
You need to find the point that is on both lines by solve the system of equations.
You can o this using substitution.
y= 1/2x-7
y=-2x+3
Because y = 1/2x-7 you can put (1/2x-7) for the y in the second equation
1/2 x - 7 = -2x+3 Add 2x to both sides.
1/2x + 2x -7 = 3 Add 7 to both sides
1/2x+4/2x = 3+7
5/2x = 10 Multiply both sides by 2/5
x=10*2/5
x=4

Now find y by substituting 4 for x in one of the equations.
y=-2(4)+3
y=-8+5
y= -5
So the point on both lines is (4,-5)

And you can now find the distance between point (-1,5) and (4,-5) using the Pythagorean Theorem .

I hope that helps make it click for you.

Then you need to find 