The distance between the two points

Question

P of open paren x sub 1 comma y sub 1 close paren

and

R of open paren x sub 2 comma y sub 2 close paren

is

d(P, R) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

. Distance formula. Find the closest point on the line

y equals 2 x

to the point

open paren 1 comma 7 close paren

. (Hint: Every point on

y equals 2 x

has the form

open paren x comma 2 x close paren

, and the closest point has the minimum distance.)

Accepted Answer

Hey, everyone in this problem, we're asked to find the coordinates of the closest point on the line Y equals three X to the 30.2 comma 14. We're told to use the distance formula. We're given four answer choices. Option A 4.4 comma 13.2, option B four, comma 13, option C four comma negative 13 and option D negative 4.4 comma 13.2. Now let's think about our line Y equals three X. OK. If we wanna write a point on the line, we can write this as X comma three X. OK? So our X value in our point is just X, we're gonna substitute in X and when we substitute in X, our Y value is three X. OK? So we have X comma three X for the X and corresponding Y value for any particular point on this line. All right. So we've written the coordinates of a general point on this line. So the distance now between the line and the 0.2 comma 14 is actually gonna be the distance between the two points. OK? So the distance between Y equals three X and two comma 14 is the distance between this point X comma three X and two comma Fort. OK. So we've turned this into the distance from a point to a line to the distance between two points. Now we know that the distance between two points D OK, using our distance formula is going to be the square root of X two minus X one squared plus Y two minus Y one squared. OK. So we take the difference between the X values of the points, the difference between the Y values of the points square them add them together, take the square root. So for our problem, we get that the distance is equal to the square root of X minus two. OK? We have the X value from the point representing the line minus the X value of the point. We were given square it. And then we're gonna add the Y value of the point representing the line three, X minus the Y value of the point we were given 14th. And again, that's squared. Now we want the closest point on the line to that point. OK. So the closest point online means that we want the smallest distance, OK? We're looking at the distance between A P A point on the line to a particular point, we want the closest point. So we want the smallest distance. Now we have this square root which is, can be a little bit not nice to deal with when you're trying to find a minimum. OK. It's a lot easier when we're working with just a polynomial. What you'll notice though is that the minimum distance D will occur at the scene X value as the minimum, the squared value. OK. So if D is its absolute minimum, then D squared will also be at its minimum. OK? Because it's just the square of the value. When we take the square, it increases with distant. All right. So now instead of focusing on D, we can focus on D squared, which is gonna help us get rid of that square root. So D squared is equal to X minus two squared plus three, X minus 14 squared. Now we can expand this, we want to write this in expanded form. So this is gonna be X squared minus four X plus four from the first factor. And from the second, we have nine X squared minus X plus 196. And you can expand that. However, you like to expand them whether that be with the foil method, any way that you like to do it. And we're just gonna simplify and collect light terms X squared plus nine X squared. Be it 10 X squared negative four, X minus 84 X gives us minus 88 X four plus 196. So we have plus 200 at the end of our D squared function. So we have the distance script and we've said that the distance squared is going to have a minimum at that same X value that the distance has a minimum. OK, something squared is minimized when that something is minimized. Now this is a positive leading coefficient. So this parabola is going to open upwards. So the minimum of a parabola that opens upwards is at the vertex. Let's write this down minimum of a parabola that opens upwards is at the vertex. Huh Now we're called it the X coordinate of the vertex is given by X is equal to negative B divided by two by, in our equation A is gonna be the coefficient corresponding with the X squared term. So that's 10 and B is gonna be the coefficient corresponding with the X term which is negative 88. And so the X value of our vertex is going to be at negative negative 88 divided by two multiplied by 10. So this is going to occur at 88 divided by 20. And this gives us an X value of 4.4. OK. So we found the X value of the closest point on that line. OK. Remember what we're trying to find, we're trying to find the coordinates of the closest point on the line. So we found the X value. Now we need to find the corresponding Y value and the corresponding one value is gonna be given by that equation of the line Y is equal to three X. OK. So it's gonna be equal to three, multiplied by 4.4, which gives us 13.2, right? And so the closest point line to the point we were given is going to be given by 4.4 comma 13.2. OK. We used our distance formula, OK, looking at the distance between a point on the line and the point we were given, OK, we use our distance formula to find the minimum where that minimum occurs. We found the X value where we have the minimum distance A K A, the closest point. And then we substituted that X value back into the equation of our line to find the corresponding widely. OK. And so we get the closest point on the line to the 0.0.2 comma is 4.4 comma 13.2 which corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Distance Formula

Equation of a Line

Minimizing Distance Using Calculus or Algebra

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