Math

# Find the distance between the following pairs of points(2,3) , (4,1)(-5,7) ,(-1,3)(a,b) ,(-a,-b)(c,o) ,(o,-c)(4,5) ,(-3,2)(at²,2at),(at²,2at²)​

Given :-

(2,3) , (4,1)

(-5,7) ,(-1,3)

(a,b) ,(-a,-b)

(c,o) ,(o,-c)

(4,5) ,(-3,2)

(at²,2at),(at²,2at²)

To find :-

Find the distance between the following pairs of points?

Solution:-

i)Given points are (2,3) and (4,1)

Let (x1, y1) = (2,3) => x1 = 2 and y1 = 3

Let (x2, y2) = (4,1) => x2 = 4 and y2 = 1

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

On substituting these values in the above formula then

=> Distance =√[(4-2)²+(1-3)²] units

=> Distance =√[2²+(-2)²]

=> Distance = √(4+4)

=> Distance = √8

=> Distance = √(2×2×2)

=> Distance = 2√2 units

Therefore, Distance between two points (2,3) and (4,1) is 2√2 units

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ii)Given points are (-5,7) and (-1,3)

Let (x1, y1) = (-5,7) => x1 = -5 and y1 = 7

Let (x2, y2) = (4,1) => x2 = -1 and y2 = 3

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

On substituting these values in the above formula then

=> Distance =√[(-1-(-5))²+(3-7)²] units

=> Distance =√[(-1+5)²+(-4)²]

=> Distance = √(4²+(-4)²)

=> Distance = √(16+16)

=> Distance = √(2×16)

=> Distance = 4√2 units

Therefore, Distance between two points (-5,7) and (-1,3) is 4√2 units

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iii)Given points are (a,b) and (-a,-b)

Let (x1, y1) = (a,b) => x1 = a and y1 = b

Let (x2, y2) = (-a,-b) => x2 = -a and y2 = -b

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

On substituting these values in the above formula then

=> Distance =√[(-a-a)²+(-b-b)²] units

=> Distance =√[(-2a)²+(-2b)²]

=> Distance = √(4a²+4b²)

=> Distance = √[4(a²+b²)]

=> Distance = 2√(a²+b²) units

Therefore, Distance between two points (a,b) and (-a,-b) is 2√(a²+b²) units

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iv)Given points are (c,o) and (o,-c)

Let (x1, y1) = (c,o) => x1 = c and y1 = o

Let (x2, y2) = (o,-c) => x2 = o and y2 = -c

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

On substituting these values in the above formula then

=> Distance =√[(o-c)²+(-c-o)²] units

=> Distance =√[o²+c²-2oc+c²+2oc+o²]

=> Distance = √(o²+c²+o²+c²)

=> Distance = √(2c²+2o²)

=> Distance = √[2(c²+o²)]units

Therefore, Distance between two points (c,o) and (o,-c) is √[2(c²+o²)] units

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v)Given points are (4,5) and (-3,2)

Let (x1, y1) = (4,5) => x1 = 4 and y1 = 5

Let (x2, y2) = (-3,2) => x2 = -3 and y2 = 2

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

On substituting these values in the above formula then

=> Distance =√[(-3-4)²+(2-5)²] units

=> Distance =√[(-7)²+(-3)²]

=> Distance = √(49+9)

=> Distance = √58 units

Therefore, Distance between two points (4,5) and (-3,2) is √58 units

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vi)Given points are (at²,2at) and (at²,2at²)

Let (x1, y1) = (at²,2at) => x1 = at² and y1 = 2at

Let (x2, y2) = (at²,2at²) => x2 = at² and y2 = 2at²

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

On substituting these values in the above formula then

=> Distance =√[(at²-at²)²+(2at²-2at)²] units

=> Distance =√[0+(2at²-2at)²]

=> Distance = √(2at²-2at)²

=> Distance = 2at²-2at or

=> Distance = 2at(t-1) units

Therefore, Distance between two points (at²,2at) and (at²,2at²) is (2at²-2at) units or 2at(t-1) units

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1) 2√2 units

2)4√2 units

3)2√(a²+b²) units

4)√[2(c²+o²)] units

5)√58 units

6)(2at²-2at) units or 2at(t-1) units

Used formulae:-

The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units

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