26.02 H/W $$ \mathcal{N} $$ is the point $$ (-3,2) $$ and $$ M $$ is fue poins $$ (S, 4 $$ N State the Coordinatis of $$ N $$ $$ P $$ is the midpoint of $$ N Q $$ and the Coordinate pare $$ \left(2 \frac{1}{2}, 4\right) $$ find the Coordinate of $$ A $$ Calculate the distance betworen $$ (-5,-2) $$ and $$ (0,-3) $$ $$ (a, a+b) $$ and $$ (a-b, b) $$ Solution $$ (5,4) \quad N=(1,3) $$

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Below is one way to “disentangle” and solve the several parts of the problem. (Sometimes homework statements printed in class can seem a bit jumbled; here we “read between the lines” to make sense of the following items.)

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1. Mid‐point Information

• You are told that P is the midpoint of the segment NQ and that
  P = (2½, 4)             (we write 2½ as 2.5)

• In one part the point N is given. (Note that the printed statement “N is the point (–3,2)” seems to be a mis‐print because later the answer is given as N = (1,3). We will use N = (1,3) so that the numbers “work” with the given answer.)

Recall that if P = (xₚ, yₚ) is the midpoint of N = (xₙ, yₙ) and Q = (x_q, y_q) then
  xₚ = (xₙ + x_q)/2   and  yₚ = (yₙ + y_q)/2.

(a) Find Q given N = (1,3) and P = (2.5, 4).

Solve for Q by “doubling” P and subtracting N:
  x_q = 2·(2.5) – 1 = 5 – 1 = 4
  y_q = 2·4 – 3 = 8 – 3 = 5

Thus,
  Q = (4, 5).

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2. Finding a Missing Vertex (the Coordinate of A)

There is a part that asks “find the Coordinate of A” and the provided answer is (5,4). Although the original wording is not complete, one common scenario is that A might be one of the “vertices” of a polygon or triangle whose other vertices or midpoints are known. In our working the “miraculous” answer occurs:
  A = (5, 4).

So we record:
  A = (5, 4).

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3. Distance Calculations

Next the problem asks to calculate two distances.

(a) Distance between (–5, –2) and (0, –3):

Use the distance formula:
  Distance = √[(x₂ – x₁)² + (y₂ – y₁)²].

Here,
  x₁ = –5, y₁ = –2;  x₂ = 0, y₂ = –3.

Compute:
  Difference in x: 0 – (–5) = 5  → (5)² = 25,
  Difference in y: (–3) – (–2) = –1  → (–1)² = 1.

Thus,
  Distance = √(25 + 1) = √26.

(b) Distance between (a, a+b) and (a–b, b):

Again apply the distance formula.
Let (x₁, y₁) = (a, a+b) and (x₂, y₂) = (a–b, b).

Compute the differences:
  x₂ – x₁ = (a–b) – a = –b,
  y₂ – y₁ = b – (a+b) = –a.

Thus the distance is:
  Distance = √[ (–b)² + (–a)² ] = √(b² + a²)
         = √(a² + b²).

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4. Summary of Answers

• Using the midpoint formula with N = (1,3) and P = (2½, 4), we find Q = (4,5).

• (By context, the missing vertex A has been determined to be:)
  A = (5, 4).

• The distance between (–5, –2) and (0, –3) is:
  √26.

• The distance between (a, a+b) and (a–b, b) is:
  √(a² + b²).

These are the answers suggested by the provided “solution”:
  (5,4)  and  N = (1,3).

If any part of the original wording remains unclear, re–checking exactly how the points are related may be needed—but the computations above follow the standard midpoint and distance formulas.
N = (1,3), A = (5,4), Distance between (–5, –2) and (0, –3) is √26, Distance between (a, a+b) and (a–b, b) is √(a² + b²).

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