Question

Determine the points in

(i) xy-plane

(ii) yz-plane and

(iii) zx-plane which are equidistant from the points A(1,-1,0), B(2, 1, 2) and C(3, 2, -1)

Answer 1

Let the point on xy-plane be P(x, y, 0) Now P is equidistance from A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1) So, AP = BP = CP

Now,

$(AP{)}^2=(x-1{)}^2+(y+1{)}^2+(0-0{)}^2$

$(BP{)}^2=(x-2{)}^2+(y-1{)}^2+(0-2{)}^2$

$(CP{)}^2=(x-3{)}^2+(y-2{)}^2+(0+1{)}^2$

$(AP{)}^2=(BP{)}^2\Rightarrow (x-1{)}^2+(y+1{)}^2$

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

$\Rightarrow x^2+1-2+y^2+1+2y+z^2$

$=x^2+4-4x+y^2+1-2y+4$

$\Rightarrow 2x+4y=7...\left(i\right)$

$(BP{)}^2=(CP{)}^2\Rightarrow (x-2{)}^2+(y-1{)}^2$

$+4=(x-3{)}^2+(y-2{)}^2+1$

$\Rightarrow x^2+4-4x+y^2+1-2y+z^2$

$+4=x^2+9-6x+y^2+4-4y+1$

$\Rightarrow 2x+2y=5...\left(ii\right)$

$(AP{)}^2=(CP{)}^2\Rightarrow (x-1{)}^2+(y+1{)}^2$

$=(x-3{)}^2+(y-2{)}^2+1$

$\Rightarrow x^2+1$

$-2x+y^2+1+2y=x^2$

$+9-6x+y^2+4-4y+1$

$\Rightarrow 4x+6y=12...\left(iii\right)$

Solving equation (i) and (ii) we get

y = 1, x = $\frac{3}{2}$

Put x and y in equation (iii)

4$\frac{3}{2}$ + 6(1) = 12

12 = 12

So, the required point is $\frac{3}{2},1,0$

Let Q (0, y, z) be the required point,

So,

$(AQ{)}^2=(BQ{)}^2\Rightarrow (0-1{)}^2+(y+1{)}^2$

$+(z-0{)}^2=(0-2)+(y-1{)}^2+(z-2{)}^2$

$\Rightarrow 1+y^2+1+2y+z^2=4+y^2+1-2y+z^2+4-42$

$\Rightarrow 4y+4z=7$

$(BQ{)}^2=(CQ{)}^2\Rightarrow (0-z^2)+(y-1{)}^2+(z-2{)}^2=(0-3{)}^2+(y-2{)}^2+(2+1{)}^2$

$\Rightarrow 4+y^2+1-2y+z^2+4-4z=9+y^2+4-4y+z^2+1+2z$

$\Rightarrow 2y-6z=5...\left(ii\right)$

$(AQ{)}^2=(CQ{)}^2\Rightarrow (0-1{)}^2+(y+1{)}^2+(z-0{)}^2=(0-3{)}^2+(y-2{)}^2+(z+1{)}^2$

$\Rightarrow 1+y^2+2y+1+z^2=9+y^2-4y+4+z^2+1+2z$

$\Rightarrow 6y-2z=12...\left(iii\right)$

$\text{Solving equation (i) and (ii), we get}$

$z=\frac{-3}{16}andy=\frac{31}{16}$

$\text{Put the value of y and z is equation}\left(iii\right)$

$6y-2z=12=12$

$6\left(\frac{31}{16}\right)-2\left(\frac{-3}{16}\right)=12$

$\frac{186}{16}+\frac{6}{16}=12$

$\frac{192}{16}=12$

12=12

LHS=RHS

So,

$y=\frac{31}{16},z=\frac{13}{16}$

$\text{Required point}=(0,\frac{31}{16},\frac{-3}{16})$

(iii) Let R (x, 0, z) be the required point.

So

$(AR{)}^2=(BR{)}^2\Rightarrow (1-x{)}^2+(-1-0{)}^2+(0-z{)}^2=(2-x)+(1-0{)}^2+(2-z{)}^2$

$\Rightarrow 1+x^2-2x+1+z^2=4+x^2-4x+1+4+z^2-4z$

$\Rightarrow 2x+4z=7...\left(1\right)$

$(BR{)}^2=(CR{)}^2\Rightarrow (z-z{)}^2+(1-0{)}^2+(2-z{)}^2=(3-x{)}^2+(2-0{)}^2+(-1-z{)}^2$

$\Rightarrow 4+x^2-4x+4+z^2-4z=9+x^2-6x+4+1+z^2+2z$

$\Rightarrow 2x-6z=5...\left(2\right)$

$(AR{)}^2=(CR{)}^2\Rightarrow (1-x{)}^2+(1-0{)}^2+(0-z{)}^2=(3-x{)}^2+(2-0{)}^2+(-1-z{)}^2$

$\Rightarrow 1+x^2-2x+1+z^2=9+6x+4+1+z^2+2z$

$\Rightarrow 4x-2z=12...\left(3\right)$

$\text{Solving equation (1) and (2), we get}$

$z=\frac{1}{5},x=\frac{31}{10}$

$\text{Put the value of x and z is equation}\left(3\right)$

$4x-2z=12$

$\text{Put the value of x and z is equation}\left(3\right)$

$4x-2z=12$

$4\left(\frac{31}{10}\right)-2\left(\frac{1}{5}\right)=12$

$\frac{124}{10}-\frac{2}{10}=12$

$\frac{120}{10}=12$

=12=12

LHS = RHS

So,

$x=\frac{31}{10},z=\frac{1}{5}$

$\text{Required point}=(\frac{31}{10},0,\frac{1}{5})$