Find the relation between x and y such that the point (x, y) is equidistant from
the points (2, 4) and (-1,3).
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-5 and
Answers
Step-by-step explanation:
Given:-
The points (2, 4) and (-1,3).
To find:-
Find the relation between x and y such that the point (x, y) is equidistant from the points (2, 4) and (-1,3).?
Solution:-
Given points are (2, 4) and (-1,3).
Let A = (2, 4)
Let B = (-1,3)
P=(x,y)
A_________P_________B
Given that P(x,y) is equidistant from the points A and B
=> AP = PB
Distance between A and P = Distance between P and B
I) Distance between A and P:-
Let (x1, y1)=A(2,4)=>x1=2 and y1=4
Let (x2, y2)=P(x,y)=x2=x and y2=y
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units
=> AP= √[(x-2)^2+(y-4)^2]
=>AP =√[x^2-4x+4+y^2-8y+16]
=>AP =√(x^2-4x+y^2-8y+20) units----------(1)
ii)Distance between Pand B:-
Let (x1, y1)=P(x,y)=>x1=x and y1=y
Let (x2, y2)=B(-1,3)=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
=> PB= √[(-1-x)^2+(3-y)^2]
=>PB =√[1+x^2+2x+9+y^2-6y]
=>PB =√(x^2+2x+y^2-6y+10) units----------(2)
Given that
AP = PB
=>√(x^2-4x+y^2-8y+20) = √(x^2+2x+y^2-6y+10)
In squaring both sides
[√(x^2-4x+y^2-8y+20)]^2=[√(x^2+2x+y^2-6y+10)]^2
=>x^2-4x+y^2-8y+20 = x^2+2x+y^2-6y+10
=>x^2-4x+y^2-8y+20 -x^2-2x-y^2+6y-10=0
=>-4x-8y+20-2x+6y-10=0
=>-6x-2y+10=0
=>-2(3x+y-5)=0
=>3x+y-5 = 0/-2
=>3x+y-5=0
Answer:-
The required relation for the given problem is 3x+y-5=0
Used formula:-
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units