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3 parts
a
5
b

13\sqrt{13}

c

252\sqrt{5}

Step-by-step solution

10 steps

The formula for the distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by:

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

Part a· 3 steps

Find the length of the line joining $(1,2)$ and $(4,6)$.

  1. Identify the coordinates.

    Let (x1,y1)=(1,2)(x_1, y_1) = (1,2) and (x2,y2)=(4,6)(x_2, y_2) = (4,6).

  2. Substitute the coordinates into the distance formula.

    d=(4−1)2+(6−2)2d = \sqrt{(4 - 1)^2 + (6 - 2)^2}
  3. Simplify the expression.

    d=(3)2+(4)2d = \sqrt{(3)^2 + (4)^2} d=9+16d = \sqrt{9 + 16} d=25d = \sqrt{25} d=5d = \boxed{5}

Part b

13\sqrt{13}

Find the length of the line joining (−1,4)(-1,4) and (2,6)(2,6).

  1. Identify the coordinates.

    Let (x1,y1)=(−1,4)(x_1, y_1) = (-1,4) and (x2,y2)=(2,6)(x_2, y_2) = (2,6).

  2. Substitute the coordinates into the distance formula.

    d=(2−(−1))2+(6−4)2d = \sqrt{(2 - (-1))^2 + (6 - 4)^2}
  3. Simplify the expression.

    d=(2+1)2+(2)2d = \sqrt{(2 + 1)^2 + (2)^2} d=(3)2+(2)2d = \sqrt{(3)^2 + (2)^2} d=9+4d = \sqrt{9 + 4} d=13d = \boxed{\sqrt{13}}

Part c

252\sqrt{5}

Find the length of the line joining (−1,−4)(-1,-4) and (3,−2)(3,-2).

  1. Identify the coordinates.

    Let (x1,y1)=(−1,−4)(x_1, y_1) = (-1,-4) and (x2,y2)=(3,−2)(x_2, y_2) = (3,-2).

  2. Substitute the coordinates into the distance formula.

    d=(3−(−1))2+(−2−(−4))2d = \sqrt{(3 - (-1))^2 + (-2 - (-4))^2}
  3. Simplify the expression.

    d=(3+1)2+(−2+4)2d = \sqrt{(3 + 1)^2 + (-2 + 4)^2} d=(4)2+(2)2d = \sqrt{(4)^2 + (2)^2} d=16+4d = \sqrt{16 + 4} d=20d = \sqrt{20}
  4. Simplify the radical.

    d=4×5d = \sqrt{4 \times 5} d=25d = \boxed{2\sqrt{5}}
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