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

Mar 27, 2017

The equations are not the same.

#### Explanation:

Find the midpoint of YZ using the midpoint formula $\left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right) = \left(\frac{- 5 + - 5}{2} , \frac{0 + - 6}{2}\right) = \left(- 5 , - 3\right)$

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

The median from X goes from $\left(- 3 , 2\right)$ to the midpoint of YZ $\left(- 5 , - 3\right)$:
${m}_{\text{median}} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{- 3 - 2}{- 5 - - 3} = \frac{- 5}{- 2} = \frac{5}{2}$

Equation of the median $y = m x + b$:
$y = \frac{5}{2} x + b$

Use either the midpoint or point X to find b:
$2 = \frac{5}{2} \cdot - \frac{3}{1} + b$
$2 = - \frac{15}{2} + b$
$\frac{4}{2} + \frac{15}{2} = \frac{19}{2} = b$

Equation of the median: $y = \frac{5}{2} x + \frac{19}{2}$

Mar 27, 2017

Perpendicular bisector of $Y Z$ is not the same line as the median from $X$.

#### Explanation:

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

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

As midpoint of $Y Z$ is $P \left(\frac{- 5 - 5}{2} , \frac{0 + 6}{2}\right)$ i.e. $\left(- 5 , - 3\right)$,

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

and median from $X \left(- 3 , 2\right)$ is $X P$ and its equation is

$\frac{y - 2}{3 - 2} = \frac{x - \left(- 3\right)}{- 5 - \left(- 3\right)}$

or $\frac{y - 2}{1} = \frac{x + 3}{- 2}$

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

or $x + 2 y - 1 = 0$

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

Obviously perpendicular bisector of $Y Z$ is not the same line as the median from $X$.