What’s is the measure of
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
The equation of a line is y = mx + b
Where:
- m is the slope
- b is the y-intercept
First, let's find what m is, the slope of the line.
Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.
Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.
Now, just plug the numbers into the formula for m above, like this:
So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:
Now, what about b, the y-intercept?
To find b, think about what your (x,y) points mean:
- (-3,1). When x of the line is -3, y of the line must be 1.
- (2,-1). When x of the line is 2, y of the line must be -1.
Now, look at our line's equation so far: .
is what we want, the -
is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).
So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!
You can use either (x,y) point you want. The answer will be the same:
- (-3,1). y = mx + b or
, or solving for b:
.
.
- (2,-1). y = mx + b or
, or solving for b:
.
See! In both cases, we got the same value for b. And this completes our problem.
The equation of the line that passes through the points (-3,1) and (2,-1) is
Let 
be an arbitrary point on the surface. The distance between 
and the given point 
is given by the function
Note that
and 
attain their extrema, if they have any, at the same values of 
. This allows us to consider the modified distance function,
So now you're minimizing
subject to the constraint 
. This is a perfect candidate for applying the method of Lagrange multipliers.
The Lagrangian in this case would be
which has partial derivatives
Setting all four equation equal to 0, you find from the third equation that either
or 
. In the first case, you arrive at a possible critical point of 
. In the second, plugging into the first two equations gives
and plugging these into the last equation gives
So you have three potential points to check:, 
, and 
. Evaluating either distance function (I use 
), you find that
So the two points on the surface closest to the point are 
.
Note that
So now you're minimizing
The Lagrangian in this case would be
which has partial derivatives
Setting all four equation equal to 0, you find from the third equation that either
and plugging these into the last equation gives
So you have three potential points to check:
So the two points on the surface