Question

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 ?

Answer 1

Answer:

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

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

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

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

$d=\sqrt{52}\ units$

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

Answer 2

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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