If the distance from P to the points (1,2), (0,-1)are in the ratio 2:1, then the locus of P is About the author Quinn
Step-by-step explanation: Given :– The distance from P to the points (1,2), (0,-1) are in the ratio 2:1. To find :– Find the locus of P? Solution :– Let the coordinates of the point P be (x,y) Given points = (1,2), (0,-1) Let A =(1,2) Let B = (0,-1) The ratio of the distances from the points to P = 2:1 PA : PB = 2:1 PA/PB = 2/1 ——-(1) Finding the distance between P and A :– Let (x1, y1)=P(x,y)=>x1=x and y1=y Let (x2, y2)=A(1,2)=>x2=1 and y2=2 We know that The distance between two points ( x1 , y1 )and ( x2, y2 ) is √[(x2-x1)^2+(y2-y1)^2] units => PA = √[(1-x)^2+(2-y)^2] => PA=√(1^2-2(1)(x)+x^2+2^2-2(2)(y)+y^2) Since (a-b)^2 = a^2 -2ab +b^2 => PA =√(1-2x+x^2+4-4y+y^2) => PA = √(x^2+y^2-2x-4y+5) units —–(2) Finding the distance between P and B :- Let (x1, y1)=P(x,y)=>x1=x and y1=y Let (x2, y2)=B(0,-1)=>x2=0 and y2= -1 We know that The distance between two points ( x1 , y1 )and ( x2, y2 ) is √[(x2-x1)^2+(y2-y1)^2] units => PB = √[(0-x)^2+(-1-y)^2] => PB=√[(-x)^2+(-(1+y))^2)] => PB = √[(x^2+(1+y)^2] => PB = √(x^2+1^2+2(1)(y)+(y^2) Since (a+b)^2 = a^2 +2ab +b^2 => PB =√(x^2+1+2y+y^2) units ——-(3) Now PA/PB = 2/1 [√(x^2+y^2-2x-4y+5)]/[√(x^2+1+2y+y^2) ] = 2/1 On squaring both sides => [x^2+y^2-2x-4y+5)]/[(x^2+1+2y+y^2)] = 4/1 On applying cross multiplication then => 4(x^2+1+2y+y^2)=x^2+y^2-2x-4y+5 =>4x^2+4+8y+4y^2=x^2+y^2-2x-4y+5 => 4x^2+4+8y+4y^2-x^2-y^2+2x+4y-5 = 0 =>(4x^2-x^2)+(4y^2-y^2)+2x+(8y+4y)+(4-5) = 0 => 3x^2+3y^2+2x+12y+1 = 0 Answer:– Locus of the given point P for the given problem is 3x^2+3y^2+2x+12y+1 = 0 Used formulae:– The distance between two points ( x1 , y1 )and ( x2, y2 ) is √[(x2-x1)^2+(y2-y1)^2] units (a-b)^2 = a^2 -2ab +b^2 (a+b)^2 = a^2 +2ab +b^2 Reply
Step-by-step explanation:
Given :–
The distance from P to the points (1,2), (0,-1) are in the ratio 2:1.
To find :–
Find the locus of P?
Solution :–
Let the coordinates of the point P be (x,y)
Given points = (1,2), (0,-1)
Let A =(1,2)
Let B = (0,-1)
The ratio of the distances from the points to P = 2:1
PA : PB = 2:1
PA/PB = 2/1 ——-(1)
Finding the distance between P and A :–
Let (x1, y1)=P(x,y)=>x1=x and y1=y
Let (x2, y2)=A(1,2)=>x2=1 and y2=2
We know that
The distance between two points ( x1 , y1 )and ( x2, y2 ) is √[(x2-x1)^2+(y2-y1)^2] units
=> PA = √[(1-x)^2+(2-y)^2]
=> PA=√(1^2-2(1)(x)+x^2+2^2-2(2)(y)+y^2)
Since (a-b)^2 = a^2 -2ab +b^2
=> PA =√(1-2x+x^2+4-4y+y^2)
=> PA = √(x^2+y^2-2x-4y+5) units —–(2)
Finding the distance between P and B :-
Let (x1, y1)=P(x,y)=>x1=x and y1=y
Let (x2, y2)=B(0,-1)=>x2=0 and y2= -1
We know that
The distance between two points ( x1 , y1 )and ( x2, y2 ) is √[(x2-x1)^2+(y2-y1)^2] units
=> PB = √[(0-x)^2+(-1-y)^2]
=> PB=√[(-x)^2+(-(1+y))^2)]
=> PB = √[(x^2+(1+y)^2]
=> PB = √(x^2+1^2+2(1)(y)+(y^2)
Since (a+b)^2 = a^2 +2ab +b^2
=> PB =√(x^2+1+2y+y^2) units ——-(3)
Now
PA/PB = 2/1
[√(x^2+y^2-2x-4y+5)]/[√(x^2+1+2y+y^2) ] = 2/1
On squaring both sides
=> [x^2+y^2-2x-4y+5)]/[(x^2+1+2y+y^2)] = 4/1
On applying cross multiplication then
=> 4(x^2+1+2y+y^2)=x^2+y^2-2x-4y+5
=>4x^2+4+8y+4y^2=x^2+y^2-2x-4y+5
=> 4x^2+4+8y+4y^2-x^2-y^2+2x+4y-5 = 0
=>(4x^2-x^2)+(4y^2-y^2)+2x+(8y+4y)+(4-5) = 0
=> 3x^2+3y^2+2x+12y+1 = 0
Answer:–
Locus of the given point P for the given problem is 3x^2+3y^2+2x+12y+1 = 0
Used formulae:–
The distance between two points ( x1 , y1 )and ( x2, y2 ) is √[(x2-x1)^2+(y2-y1)^2] units