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Calculate the distance between the following points: (10, 8) and (4, 8.4)

Mathematics

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1. Calculate the distance between $(3,1)$ and $(7,6)$

  1. Identify the coordinates of the two points.

    Let (x1,y1)=(3,1)(x_1, y_1) = (3, 1) and (x2,y2)=(7,6)(x_2, y_2) = (7, 6).

  2. Apply the distance formula.

    The distance DD between two points is given by D=(x2−x1)2+(y2−y1)2D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

    D=(7−3)2+(6−1)2D = \sqrt{(7 - 3)^2 + (6 - 1)^2}
  3. Simplify the expression.

    D=(4)2+(5)2D = \sqrt{(4)^2 + (5)^2} D=16+25D = \sqrt{16 + 25} D=41D = \sqrt{41}
  4. Calculate the numerical value and round if necessary.

    D≈6.403D \approx 6.403

    The distance is 6.4 units\boxed{6.4 \text{ units}}.

    2. Calculate the distance between (10,8)(10,8) and (4,1)(4,1)

  5. Identify the coordinates of the two points.

    Let (x1,y1)=(10,8)(x_1, y_1) = (10, 8) and (x2,y2)=(4,1)(x_2, y_2) = (4, 1).

  6. Apply the distance formula.

    The distance DD between two points is given by D=(x2−x1)2+(y2−y1)2D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

    D=(4−10)2+(1−8)2D = \sqrt{(4 - 10)^2 + (1 - 8)^2}
  7. Simplify the expression.

    D=(−6)2+(−7)2D = \sqrt{(-6)^2 + (-7)^2} D=36+49D = \sqrt{36 + 49} D=85D = \sqrt{85}
  8. Calculate the numerical value and round if necessary.

    D≈9.219D \approx 9.219

    The distance is 9.2 units\boxed{9.2 \text{ units}}.

    3. Calculate the length of the new road from hospital (1,1)(1,1) to town (6,5)(6,5)

  9. Identify the coordinates of the two points.

    Let (x1,y1)=(1,1)(x_1, y_1) = (1, 1) and (x2,y2)=(6,5)(x_2, y_2) = (6, 5).

  10. Apply the distance formula.

    The distance DD between two points is given by D=(x2−x1)2+(y2−y1)2D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

    D=(6−1)2+(5−1)2D = \sqrt{(6 - 1)^2 + (5 - 1)^2}
  11. Simplify the expression.

    D=(5)2+(4)2D = \sqrt{(5)^2 + (4)^2} D=25+16D = \sqrt{25 + 16} D=41D = \sqrt{41}
  12. Calculate the numerical value and round if necessary.

    D≈6.403D \approx 6.403

    The length of the new road is 6.4 units\boxed{6.4 \text{ units}}.

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