Question

The relation between x and y so that the point (x,y) is equidistant from the points (-4,-4) and (-2,4) is : (a) x +4y=3 (b) x -4y+3=0 (c) x +4y+3=0 (d) x -4y=3​

Answer 1

Answer:

Solution:

\textsf{Let the given points A(-4,4), B(-2,4) and P(x,y)}Let the given points A(-4,4), B(-2,4) and P(x,y)

\textsf{As per given data,}As per given data,

\mathsf{PA=PB}PA=PB

\mathsf{\sqrt{(x+4)^2+(y-4)^2}=\sqrt{(x+2)^2+(y-4)^2}}

(x+4)

2

+(y−4)

2

=

(x+2)

2

+(y−4)

2

\textsf{Squaring on bothsides, we get}Squaring on bothsides, we get

\mathsf{(x+4)^2+(y-4)^2=(x+2)^2+(y-4)^2}(x+4)

2

+(y−4)

2

=(x+2)

2

+(y−4)

2

\mathsf{(x+4)^2=(x+2)^2}(x+4)

2

=(x+2)

2

\mathsf{x^2+16+8x=x^2+4+4x}x

2

+16+8x=x

2

+4+4x

\mathsf{x^2+16+8x=x^2+4+4x}x

2

+16+8x=x

2

+4+4x

\implies\mathsf{4x+12=0}⟹4x+12=0

\therefore\boxed{\mathsf{x+3=0}}∴

x+3=0

\textsf{whch is the required relation}whch is the required relation