Question

In a `DeltaXYZ` having `X(-3,2)`, `Y(-5,-6)` and `Z(-5,0)`, is perpendicular bisector of `YZ` same as median from `X`?

Answer 1

The equations are not the same.

Explanation:

Find the midpoint of YZ using the midpoint formula `((x_1+x_2)/2, (y_1+y_2)/2) = ((-5+ -5)/2, (0+ -6)/2) = (-5, -3)`

Since YZ is a vertical line: `x = -5`,
The perpendicular bisector is a horizontal line: `y = -3`

The median from X goes from `(-3, 2)` to the midpoint of YZ `(-5, -3)`:
`m_"median" = (y_2 - y_1)/(x_2 - x_1) = (-3 - 2)/(-5 - -3) = (-5)/(-2) = 5/2`

Equation of the median `y = mx + b`:
`y = 5/2 x +b`

Use either the midpoint or point X to find b:
`2 = 5/2 * -3/1 +b`
`2 = -15/2 + b`
`4/2 + 15/2 = 19/2 = b`

Equation of the median: `y = 5/2 x + 19/2`

Answer 2

Perpendicular bisector of `YZ` is not the same line as the median from `X`.

Explanation:

As the abscissa (i.e. `x`-coordinate) of `Y(-5,-6)` and `Z(-5,0)` both are `-5`, `YZ` is parallel to `y`-axis

and as such its perpendicular bisector will be parallel to `x`-axis.

As midpoint of `YZ` is `P((-5-5)/2,(0+6)/2)` i.e. `(-5,-3)`,

equation of perpendicular bisector is `y+3=0`.

and median from `X(-3,2)` is `XP` and its equation is

`(y-2)/(3-2)=(x-(-3))/(-5-(-3))`

or `(y-2)/1=(x+3)/(-2)`

or `x+3=-2y+4`

or `x+2y-1=0`

Note it is different from the equation of median given in question.

Obviously perpendicular bisector of `YZ` is not the same line as the median from `X`.