Question

A(_5,2) and B(4,1) . Find the equation of the locus of points P , which is equidistant from A and B​

Answer 1

Answer:

x+y=6

Step-by-step explanation:

Lets consider a point P euidistant from point A and B with co-ordinates x and y.

according to distance formula,

the distance between any two arbitrary point C(x1,y1) and D)(x2,y2)=$\sqrt{(x1-x2)^{2} +(y1-y2)^{2} }$

since,point P is equidistant from A and B

hence,PA =PB

$\sqrt{(x-5)^{2} +(y-2)^{2}}$=$\sqrt{(x-4)^{2} +(y-1)^{2}}$

$x^{2} - 10\times x+25+y^{2}-4\times y+ 4 = x^{2} - 8\times x+16+y^{2} -2\times y+1$

$2\times x+ 2\times y=29-17$

x+y=6

Hence equation of locus of point p is given by

x+y=6

Answer 2

Answer:

x+y=6

Step-by-step explanation:

Lets consider a point P euidistant from point A and B with co-ordinates x and y.

according to distance formula,

the distance between any two arbitrary point C(x1,y1) and D)(x2,y2)=\sqrt{(x1-x2)^{2} +(y1-y2)^{2} }(x1−x2)2+(y1−y2)2

since,point P is equidistant from A and B

hence,PA =PB

\sqrt{(x-5)^{2} +(y-2)^{2}}(x−5)2+(y−2)2 =\sqrt{(x-4)^{2} +(y-1)^{2}}(x−4)2+(y−1)2

x^{2} - 10\times x+25+y^{2}-4\times y+ 4 = x^{2} - 8\times x+16+y^{2} -2\times y+1x2−10×x+25+y2−4×y+4=x2−8×x+16+y2−2×y+1

2\times x+ 2\times y=29-172×x+2×y=29−17

x+y=6

Hence equation of locus of point p is given by

x+y=6