Perpendicular Bisector of the segment joining

Question

Find the perpendicular bisector of the segment joining $A(1,2)$ $B(1,-4)$. In this question, first I found the midpoint of $[AB]$ $M(1,-1)$ Then I found $N= B-A = (0,-6)$. Then we may take $u=(6,0)$. vector equation $P= M + tu$

$P= (1,-1)+t(6,0)$ $P= (6t+1, 0t-1)$ In here, $x=6t+1 y=0t-1$ Here I wanted to make the t equal to each other, but this is not possible. I will be grateful if you could help me.

Answer 1

Think about the problem geometrically, which is always a good idea before jumping to any equation-solving.

The two points given lie on the same (vertical) line $x=1$, with midpoint $M(1,-1)$, as you gave. Hence, the perpendicular line will be parallel to the $x$-axis, which must hence be $y=-1$.

You can see this too in your set of points $P(6t+1,-1)$, which cover every possible $x$-coordinate but always have $y$-coordinate $-1$.

Answer 2

Please note the straight line going through points $A$ and $B$ is parallel to $y-$ axis which is clear from the fact that $x$ coordinate is $1$ for both points $A$ and $B$. So the line perpendicular to it will be parallel to $x$ axis which means $y$ remains constant and is not a function of $t$. In this case, $y = -1$.

So points on the perpendicular bisector is given by $(6t+1, -1)$. So your working is correct and $y$ coordinate cannot be written as a function of $t$.