Question

find the distance between the following pairs of the point
(I) (-5,2) and (7,-3)
(ii) (2,0) and (-1,4)

Answer 1

Hey friend..!! here's your answer
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(i) ( -5 , 2) and ( 7 , -3)

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

$= \sqrt{ {(7 - ( - 5))}^{2} + {( - 3 - 2)}^{2} } \\ \\ = \sqrt{144 + 25 } \\ \\ = \sqrt{169}$

13 units

(ii) (2 ,0) and (-1 ,4)

$= \sqrt{ {( - 1 - 2)}^{2} + {(4 - 0)}^{2} } \\ \\ = \sqrt{9 + 16} \\ \\ = \sqrt{25}$

5 unit

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#Hope its help

Answer 2

Answer:

The distance between (-5,2) and (7,-3) is 13 units.

The distance between (2,0) and (-1,4) is 5 units.

Step-by-step explanation:

Given the two pairs of coordinate points we have to find the distance between each pair.

Using distance formula,

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

(I) (-5,2) and (7,-3)

$D=\sqrt{(7-(-5))^2+(-3-2)^2}=\sqrt{12^2+(-5)^2}=\sqrt{169}=13units$

(ii) (2,0) and (-1,4)

$D=\sqrt{(-1-2)^2+(4-0)^2}=\sqrt{(-3)^2+(4)^2}=\sqrt{25}=5units$

The distance between (-5,2) and (7,-3) is 13 units.

The distance between (2,0) and (-1,4) is 5 units.