Main content

## Topic D: Partitioning and extending segments and parameterization of lines

Current time:0:00Total duration:5:32

# Distance between point & line

## Video transcript

Find the distance between the
point negative 2, negative 4. This point right here. And the line y is equal
to negative 1/3 x plus 2. That's this line
right over here. Now to do it, we just need
to figure out a perpendicular line to this blue line, to
y is equal to negative 1/3 x plus 2, that contains
this point right over here. So we need to figure out
the equation of this line. And then we need to figure
out where do these two lines intersect,
and then we need to find the distance
between these two points of intersection, and
we have the shortest distance between this point and
this line right over here. So the first step
is to figure out what is the slope of
this perpendicular line. Well, the slope of
a perpendicular line is going to be the
negative inverse of the slope of this blue line. So the negative
inverse of negative 1/3 is going to be positive 3. So this line right over here
is going to have a slope of 3. So it's going to
have the form y is equal to 3x plus b, where
b is its y-intercept. It looks, just
eyeballing it here, that it's going to
be pretty close to 2. But let's verify that. So to figure out what b actually
is, let's substitute this point right over here. We know that not only
is this line slope 3, but this point has to sit on it. So this point has to
satisfy this equation. So when x is negative
2, y is negative 4. Or we have negative 4 is equal
to 3 times negative 2 plus b. Let me write the
negative 2 in there. 3 times negative 2 plus b. And now we can solve for b. We get negative 4 is equal
to negative 6 plus b. Add 6 to both sides, you
get 2 is equal to b or b is equal to 2. So we were right. The y-intercept for the
second line is at 2. So we immediately
can eyeball, or we can verify, where
they both intersect. They both intersect the
y-axis at y equals 2. For both of these, when x is
equal to 0, y is equal to 2. If it wasn't so
obvious, we could set these two equations
equal to each other. We could say look,
we have 3x plus 2. We know that this is now
3x plus 2, because b is 2. When does 3x plus 2 equal
negative 1/3 x plus 2? Well, let's see. If we subtract 2
from both sides, when does 3x equal
negative 1/3 x? Well, there's a couple
of things that we could do right over here. We could add 1/3
x to both sides. And then we will
get 3 and 1/3, which is the same thing as
(10/3)x is equal to 0. And if you multiply
both sides by 3/10, you get x is equal to 0. So these two lines intersect
when x is equal to 0. For both of them, when x is
equal to 0, y is equal to 2. But you could have
eyeballed it here. You could have seen that
both of their y-intercepts, which happens when x is
equal to 0, y is equal to 2. So this point right over
here is the point 0, 2. We already know that this
point right over here is the point negative
2, negative 4. And now we just need to find
the distance between these two points. And the distance
formula really is just an application of the
Pythagorean theorem. We just need to
find the distance in the change in the y
direction and the change in the x direction. So let's do that separately. So in the y direction, what is
this distance right over here? So we went from y is equal to
negative 4 to y is equal to 2. This distance right
over here is 6. And what is this
distance right over here? Well, we go from x equals
negative 2 to x equals 0. So this distance
right over here is 2. So the distance between
these two points is really just the hypotenuse
of a right triangle that has sides 6 and 2. If we call this
distance d, we could say that the distance
squared is equal to. And all I'm really doing here is
restating the distance formula. The distance formula
tells you all this Y2 minus Y1, which is 6, squared. But that's just the
Pythagorean theorem. That's just saying 6 squared
plus x2 minus x1, which is 0 minus negative 2, which
is positive 2 squared is going to be equal to the
distance squared. But we see that's just
the Pythagorean theorem. But anyway, let's
solve for the distance. So the distance
squared is going to be equal to 36 plus 4, which is 40. And now, let's see. The distance is equal to
the square root of 40. Square root of 40
is the same thing as the square root
of 4 times 10. And so that's the same. So the distance is equal to
2 if we factor out the 4. It's the square root of 4
times square root of 10. 2 is the square root of 4. 2 square roots of 10. And we're done.