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Calculate the distance between the following pairs of points: i) (-3, 4) and (-5, 0) ii) (2, 0) and (3, 4) iii) (1/2, 1) and (1, 1 1/2)

Mathematics

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Answers

3 parts
i

252\sqrt{5}

ii

17\sqrt{17}

iii

22\frac{\sqrt{2}}{2}

Step-by-step solution

8 steps

To find the length between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), we use the distance formula:

D=(x2−x1)2+(y2−y1)2D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Part i· 3 steps

For the points (−3,4)(-3, 4) and (−5,0)(-5, 0)

Let (x1,y1)=(−3,4)(x_1, y_1) = (-3, 4) and (x2,y2)=(−5,0)(x_2, y_2) = (-5, 0).

  1. Substitute the coordinates into the distance formula.

    D=(−5−(−3))2+(0−4)2D = \sqrt{(-5 - (-3))^2 + (0 - 4)^2}

  2. Simplify the terms inside the square root.

    D=(−5+3)2+(−4)2D = \sqrt{(-5 + 3)^2 + (-4)^2} D=(−2)2+(−4)2D = \sqrt{(-2)^2 + (-4)^2} D=4+16D = \sqrt{4 + 16} D=20D = \sqrt{20}

  3. Simplify the square root.

    D=4×5=25D = \sqrt{4 \times 5} = 2\sqrt{5}

    The length is 25\boxed{2\sqrt{5}}.

Part ii

17\sqrt{17}

For the points $(2, 0)$ and $(3, 4)$

Let (x1,y1)=(2,0)(x_1, y_1) = (2, 0) and (x2,y2)=(3,4)(x_2, y_2) = (3, 4).

  1. Substitute the coordinates into the distance formula.

    D=(3−2)2+(4−0)2D = \sqrt{(3 - 2)^2 + (4 - 0)^2}

  2. Simplify the terms inside the square root.

    D=(1)2+(4)2D = \sqrt{(1)^2 + (4)^2} D=1+16D = \sqrt{1 + 16} D=17D = \sqrt{17}

    The length is 17\boxed{\sqrt{17}}.

Part iii

22\frac{\sqrt{2}}{2}

For the points (12,1)(\frac{1}{2}, 1) and (1,112)(1, 1\frac{1}{2})

First, convert 1121\frac{1}{2} to an improper fraction: 112=321\frac{1}{2} = \frac{3}{2}.
So the points are (12,1)(\frac{1}{2}, 1) and (1,32)(1, \frac{3}{2}).
Let (x1,y1)=(12,1)(x_1, y_1) = (\frac{1}{2}, 1) and (x2,y2)=(1,32)(x_2, y_2) = (1, \frac{3}{2}).

  1. Substitute the coordinates into the distance formula.

    D=(1−12)2+(32−1)2D = \sqrt{\left(1 - \frac{1}{2}\right)^2 + \left(\frac{3}{2} - 1\right)^2}

  2. Simplify the terms inside the square root.

    D=(22−12)2+(32−22)2D = \sqrt{\left(\frac{2}{2} - \frac{1}{2}\right)^2 + \left(\frac{3}{2} - \frac{2}{2}\right)^2} D=(12)2+(12)2D = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} D=14+14D = \sqrt{\frac{1}{4} + \frac{1}{4}} D=24D = \sqrt{\frac{2}{4}} D=12D = \sqrt{\frac{1}{2}}

  3. Simplify the square root.

    D=12=12D = \frac{\sqrt{1}}{\sqrt{2}} = \frac{1}{\sqrt{2}}

    To rationalize the denominator:

    D=12×22=22D = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

    The length is 22\boxed{\frac{\sqrt{2}}{2}}.

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