Question

The point $P$ is on the $y$-axis. If $P$ is equidistant from $(1,2,3)$ and $(2,3,4),$ then $P_y=$

• A. $\frac{15}{2}$
• B. $\frac{3}{2}$
• C. $1$
• D. $3$

Answer 1

The correct option is A $\frac{15}{2}$

Given: $A(1,2,3)$ and $B(2,3,4),$
Let the point $P$ be $(0,y,0)$
Since, it is equidistant from the two points $A,B$
$A{P}^2=P{B}^2$
$\Rightarrow (1+(y-2{)}^2+9)=(4+(y-3{)}^2+16)$
$\Rightarrow (10+y^2-4y+4)=(20+y^2-6y+9)$
$\Rightarrow y^2-4y+14=y^2-6y+29$
$\Rightarrow 2y=15$
$\Rightarrow y=\frac{15}{2}$