How do you find the distance between (-2,-4) and (-5,-3)?

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How do you find the distance between (-2,-4) and (-5,-3)?

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Answer 1 · 2017-08-10T11:24:12

$d = \sqrt{10}$ $d = sqrt(10)$ units

Explanation:

Use the distance formula in coordinate geometry.

$d = \sqrt{{(y_{2} - y_{1})}_{2}^{2} + {(x_{2} - x_{1})}_{2}^{2}}$ $d = sqrt((y_2 - y_1)^2+(x_2 - x_1)^2)$

Label your points with $(x_{1}, y_{1})$ $(x_1, y_1)$ and $(x_{2}, y_{2})$ $(x_2, y_2)$ .

Let's let $(- 2, - 4)$ $(-2,-4)$ be $(x_{2}, y_{2})$ $(x_2, y_2)$ , and $(- 5, - 3)$ $(-5, -3)$ be $(x_{1}, y_{1})$ $(x_1, y_1)$ . Then we substitute in the values into the equation and solve.

$d = \sqrt{{((- 4) - (- 3))}^{2} + {((- 2) - (- 5))}^{2}}$ $d = sqrt(((-4) - (-3))^2+((-2) - (-5))^2)$

$d = \sqrt{1 + 9}$ $d = sqrt(1+9)$

$d = \sqrt{10}$ $d = sqrt(10)$

Therefore, the distance is $\sqrt{10}$ $sqrt(10)$ units.

Answer 2 · 2017-08-10T13:23:59

$\sqrt{10}$ $color(magenta)(sqrt 10$ or $\pm 3.162$ $color(magenta)(+-3.162$ units to the nearest 3 decimal places

Explanation:

$:.y_2-y_1$

$:.(-4)-(-3)=-1=opposite$

$:.x_2-x_1$

$:.(-2)-(5)=3=adjacent$

Pythagoras:

$:.d^2=(-1)^2+(3)^2$

$:.d^2=1+9$

$:.d^2=10$

$:.color(magenta)(d=sqrt 10=+-3.162$ to 3 decimal places

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How do you find the distance between (-2,-4) and (-5,-3)?

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