Question

find the new point of the locus of point P such that PA²+ PB²=PC² where A(1,-2) B(-2,4) C(3,3)​

Answer 1

$\huge{\boxed{\rm{\red{x²+y²+8x+2y+7=0}}}}$

Step-by-step explanation:

Given:-

PA²+ PB²=PC² where A(1,-2) B(-2,4) C(3,3)

To find:-

find the new point of the locus of point P such that PA²+ PB²=PC² where A(1,-2) B(-2,4) C(3,3).

Solution:-

Let the point on the locus be P(x,y)

and A(1,-2) B(-2,4) C(3,3)

Distance formula=√{(x2-x1)²+(y2-y1)²} units

(x1,y1)=(x,y)=>x1=x;y1=y

(x2,y2)=(1,-2)=>x2=1;y2=-2

PA²=[√{(1-x)²+(-2-y)²}]²

=>PA²=(1-x)²+(-2-y)²

=>PA²=1+x²-2x+4+y²+4y

=>PA²=x²+y²-2x+4y+5------(1)

and

(x1,y1)=(x,y)=>x1=x;y1=y

(x2,y2)=(-2,4)=>x2=-2;y2=4

PB²=[√{(-2-x)²+(4-y)²}]²

=>PB²=(-2-x)²+(4-y)²

=>PB²=4+x²+4x+16+y²-8y

=>PB²=x²+y²+4x-8y+20------(2)

and

(x1,y1)=(x,y)=>x1=x;y1=y

(x2,y2)=(3,3)=>x2=3;y2=3

PC²=[√{(3-x)²+(3-y)²}]²

=>PC²=(3-x)²+(3-y)²

=>PC²=9+x²-6x+9+y²-6y

=>PC²=x²+y²-6x-6y+18------(3)

On adding (1)&(2)

PA²+PB²

=x²+y²-2x+4y+5+x²+y²+4x-8y+20

=>2x²+2y²+2x-4y+25-----(4)

Now we have

PA²+ PB²=PC²

=>2x²+2y²+2x-4y+25=x²+y²-6x-6y+18

= (2x²-x²)+(2y²-y²)+(2x+6x)+(6y-4y)+(25-18)=0

=>x²+y²+8x+2y+7=0

Answer:-

The point of the locus of P=x²+y²+8x+2y+7=0