Question

if the distance of p(x,y) from a (6,2) and b(-2,6) are equal . prove that y=2x..... solve it​

Answer 1

Distance of p( x,y) from (6,2)

is √( 6 -x)^2 + ( 2- y)^2

= √36 + x^2 - 12x + 4 +y^2 - 4y

= √x^2 + y^2 -12x -4y + 40

Distance of p(x,y) from (-2,6)

√( -2-x)^2 + ( 6-y)^2

= √4 + x^2 + 4x + 36 +y^2 -12y

As distances are equal

√x^2 + y^2 - 12x -4y +40 = √x^2 + y^2 +4x -12y + 40

Squaring

x^2 + y^2 -12x -4y +40 = x^2 +y^2 +4x -12y +40

-12x -4x = -12y +4y

-16x = -8y

x= y/2

y= 2x

Answer 2

$\sf\huge{\underline{\boxed{Question}}}$

If the distance of p(x,y) from a (6,2) and b(-2,6) are equal. Then find the co-ordinates of p(x,y).

$\sf\huge{\underline{\boxed{Answer!}}}$

Given: Since the point p(x,y) is at equidistant from the two points i.e, a(6,2) and b(-2,6)

we have mid point formula!

p(x,y)=($\frac{x1+x2}{2},\frac{y1+y2}{2}$)

we have!

a(6,2) b(-2,6)

(x1,y1) (x2,y2)

substituting..

(x,y)=($\frac{x1+x2}{2},\frac{y1+y2}{2}$)

x = $\frac{x1+x2}{2}$

x = $\frac{6+(-2)}{2}$

x = $\frac{6-2}{2}$

x= $\frac{4}{2}$

x = 2..==(1)

SIMILARLY,

y= $\frac{y1+y2}{2}$

= $\frac{2+6}{2}$

= $\frac{8}{2}$

y = 4==(2)

hence the co-ordinates of points p(x,y) are p(2,4)

comparing (1)&(2)

x = 2, y =4

i.e, y = 2x = 2(2) = 4..

##hence the proof!!