Math, asked by devikadeepu2004, 2 months ago

Find the relation between x and y such that the point (x, y) is equidistant from
the points (2, 4) and (-1,3).
th
-5 and​

Answers

Answered by tennetiraj86
0

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

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