The pointegral which is equidistant from A(3, 4, -1) and B(1... | MATH - SYSTEM OF CO-ORDINATES, DIRECTION COSINES & DIRECTION RATIOS, PROJECTION | SolveeIt

The point which is equidistant from A(3, 4, -1) and B(1, -2, 5) on Y-axis is

(0,1, 0)

$\left( 0 , \frac { 1 } { 3 } , 0 \right)$

$\left( 0 , - \frac { 1 } { 3 } , 0 \right)$

$\left( 0 , - \frac { 5 } { 3 } , 0 \right)$

Answer

$\left( 0 , - \frac { 1 } { 3 } , 0 \right)$

Explanation

The plane that perpendicularly bisects $\overline { \mathrm { AB } }$ is

$2 ( x - 2 ) + 6 ( y - 1 ) - 6 ( z - 2 ) = 0$ (i.e.)

$x + 3 y - 3 z + 1 = 0$

This cuts **Y-**axis at $\left( 0 , - \frac { 1 } { 3 } , 0 \right)$

Question: The point which is equidistant from A(3, 4, -1) and B(1, -2, 5) on Y-axis is...