Mathematics: To find the distance between (-2,-3) and (4,1), peggy s teacher drew a diagram and marked blue lines to represent the legs of a right triangle. peggy then used 42 and 62 to find the distance represented by the red segment. what distance did she find ?

To find the distance between (-2,-3) and (4,1), - id23413960255, lovelyheart5337

matthewt894

11.01.2022

The distance between (-2,-3) and (4,1) is $\sqrt{61}$ units

The distance between two points (x_1, y_1) and (x_2, y_2) is given by the formula:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

For the points (-2, -3) and (4, 1):

x₁ = -2, y₁ = -3, x₂ = 4 y₂ = 1

Substitute x₁ = -2, y₁ = -3, x₂ = 4 y₂ = 1 into the formula above:

$D = \sqrt{(4-(-2))^2+(1-(-3))^2}$

$D = \sqrt{(4+2)^2+(1+4)^2} \\\\D = \sqrt{6^2+5^2} \\\\D = \sqrt{36+25} \\\\D = \sqrt{61}$

The distance between (-2,-3) and (4,1) is $\sqrt{61}$ units

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briarkaltvedt

11.01.2022

$d=2\sqrt{13}\ units$

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

in this problem we have

$A(-2,-3)\\B(4,1)$

substitute the values

$d=\sqrt{(1+3)^{2}+(4+2)^{2}}$

$d=\sqrt{(4)^{2}+(6)^{2}}$

$d=\sqrt{52}\ units$

$d=2\sqrt{13}\ units$

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korirosekc

16.12.2020

I hope this helps you

h.h-2.10.h+(-10).(-10)

(h-10)^2

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