Segment mn has a midpoint p. if m has coordinates (14,-3) and p has coordinates (-8,6) what are the coordinates of n?
can you explain this to me?

The coordinates of N are (-30, 15).

Step-by-step explanation:

1. Basically what the question is saying is that there are two points (M and N) connected by a line, and that the midpoint of this line is the point P.

What we need to realise is that to find the midpoint of something, we would usually divide the length by 2. In this case, because we are working on an axis with two dimensions (x and y), the midpoint is found by calculating the distance between the points M and N in both length (x) and height (y), and dividing these values by 2.

2. If we were to take this concept and translate this into a formula, we would get:

Midpoint = ( (x1 + x2) / 2 ), (y1 + y2) / 2 ),

where (x1, y1) and (x2, y2) are coordinates of the two endpoints of the line segment.

3. In the question itself, we are given that the midpoint P has coordinates (-8, 6) and the point M has coordinates (14, -3). Our aim is to find the coordinates of N, which we will denote as (x2, y2) (note that it wouldn't matter if we had denoted it as (x1, y1), the answer would still be the same).

Substituting these values into the formula we defined above, we get:

(-8, 6) = ( (14 + x2)/2), (-3 + y2)/2 )

Here, you should be able to see that -8 is the average of the x-values of points M and N, and 6 is the average of the y-values of points M and N. Thus, we can write these out separately and solve for the missing coordinate pair (x2, y2):

(i) Finding x2:

-8 = (14 + x2)/2

-16 = 14 + x2 (Multiply both sides by 2)

-30 = x2 (Subtract 14 from both sides)

(ii) Finding y2:

6 = (-3 + y2)/2

12 = -3 + y2 (Multiply both sides by 2)

15 = y2 (Add 3 to both sides)

4. From the calculations above, we now know that x2 = -30 and y2 = 15, thus the coordinates of N are (-30, 15).

I hope that helped clear things up a little, but if you have any further questions please feel free to comment below :)

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step-by-step explanation: