Question

show that distance between (-1,3 ) (4,-2) is 5 root 2 units
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Answer 1

Answer:

let x1=-1,x2=4,y1=3,y2=-2

Step-by-step explanation:

let A and B are the points, then distance between AB square =(4-(-1)2+(-2-3)2. =(5)2+(-5)2. = 25+25. AB square =50. taking square root, AB =5 root 2

$\sqrt{?}$

Answer 2

The distance between the given points is $s=5\sqrt{2}$ units

Step-by-step explanation:

Given that the points are (-1,3 ) and (4,-2)

To show that the distance between (-1,3 ) and (4,-2) is $5\sqrt{2}$ units :

Let $(x_1,y_1)$ and $(x_2,y_2)$ be the given points (-1,3 ) and (4,-2) respectively

The distance between the two points formula is

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

• Now substitute the points in the formula we get,
• $s=\sqrt{4-(-1))^2+(-2-3)^2}$ units
• $=\sqrt{4+1)^2+(-5)^2}$
• $=\sqrt{5^2+5^2$
• $=\sqrt{25+25}$

• $=\sqrt{50}$
• $=\sqrt{25\times 2}$
• $=\sqrt{25}\times \sqrt{2}$  ( by using the property $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$ )
• $=5\times \sqrt{2}$
• $=5\sqrt{2}$ units

Therefore the distance between the given points is $s=5\sqrt{2}$ units

Hence proved.