Question

Find the coordinate of a point equidistant from (1, -6), (5, -6) and (6, -1)

• A. (2, -2)
• B. (3, -2)
• C. (3, -3)
• D. (2, -3)
• E. none of these

Answer 1

The correct option is C (3, -3)

Let $(x,y)$ be the coordinate of the point equidistant from $(1,-6)(5,-6)$ and $(6,-1)$

Now,

Distance between $(x,y)$ and $(1,-6)$ is

$d_1=\sqrt{(x-1{)}^2+(y+6{)}^2}$

$=>D_1^2=(x-1{)}^2+(y+6{)}^2$ (i)

Distance between $(x,y)$ and $(5-6)$ is

$D-2=\sqrt{(x-5{)}^2+(y+6{)}^2}$

$=>D_2^2=(x-5{)}^2+(y+6{)}^2$(ii)

Distance between $(x,y)$ and $(6,-1)$ is

$D_3=\sqrt{(x-6{)}^2+(y+1{)}^2}$

$D_3^2=(x-6{)}^2+(y+1{)}^2$ (iii)

$D_1=D_2=D_3$

Subtracting (Ii) from (i) we get,

$=>D_1^2-D_2^2=(x-1{)}^2+(y+6{)}^2-(x-5{)}^2-(y-6{)}^2$

$=>0=(x-1{)}^2-(x-5{)}^2$

$=>x^2-2x+1+(x^2-10x+25)=0$

$=>-2x+10x+1-25=0$

$=>8x=24$

$=>x=3$

Subtracting (iii) from (i) and putting $x=3$ we get,

$=>D_1^2-D_3^2=(3-1{)}^2+(y+6{)}^2-(3-6{)}^2-(y+1{)}^2$

$=>0=4-9+(y^2+12y+36)-(y^2+2y+1)$

$=>5=y^2+12y+36-y^2-2y-1$

$=>5=10y+35$

$=>10y=-30$

$=>y=-3$

$\therefore$ coordinate is $(3,-3)$