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

$\underline{\textbf{Given:}}$

$\textsf{(x,y) is equidistant from the points (-4,4) and (-2,4)}$

$\underline{\textbf{To find:}}$

$\textsf{The relation between the points}$

$\underline{\textbf{Solution:}}$

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

$\textsf{As per given data,}$

$\mathsf{PA=PB}$

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

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

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

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

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

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

$\implies\mathsf{4x+12=0}$

$\therefore\boxed{\mathsf{x+3=0}}$

$\textsf{whch is the required relation}$

$\underline{\textbf{Find more:}}$

Find the locus of point P if PA=PB. Coordinates of. A(4,0)

and B(-4,0)

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