Find the distance between the following pair of points:( - 2, - 3) and (3, 2)
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distance of (-2 ,-3) and (3,2)
let be x₁=-2 ,x₂=3 ,y₁=-3 ,y₂=2
=√ (x₁-x₂)²+(y₁-y₂)²
=√(-2-3)²+(-3-2)²
=√25+25
=√50
=5√2
let be x₁=-2 ,x₂=3 ,y₁=-3 ,y₂=2
=√ (x₁-x₂)²+(y₁-y₂)²
=√(-2-3)²+(-3-2)²
=√25+25
=√50
=5√2
CUTEBARBIE:
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Answered by
0
5√2 Units the distance between the two points.
Here is how I got it.
(x2-x1)*2 + (y2-y1)*2 (*2 means the whole square)
Distance =√(x2−x1)2+(y2−y1)2
Application of this formula leads to:
√(3-2)*2 + (2+3)*2
√5*2 + 5*2= √25+25=50
5√2 units is the answer!
Mark as brainyest :)
Here is how I got it.
(x2-x1)*2 + (y2-y1)*2 (*2 means the whole square)
Distance =√(x2−x1)2+(y2−y1)2
Application of this formula leads to:
√(3-2)*2 + (2+3)*2
√5*2 + 5*2= √25+25=50
5√2 units is the answer!
Mark as brainyest :)
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