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)
Answers
Step-by-step explanation:
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
_____________________________
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
_____________________________
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
_____________________________
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
____________________________
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
_____________________________
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
____________________________
Answer :-
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