# Thread: coordinate geometry question 2

1. ## coordinate geometry question 2

find the value of "K" if the point P(0,2) lies on the perpendicular bisector of the segment joining the points (3,K) and (K,5).

2. Originally Posted by nitinnitin2
find the value of "K" if the point P(0,2) lies on the perpendicular bisector of the segment joining the points R(3,K) and S(K,5).
1. Calculate the slope of RS: $m_{RS}=\dfrac{5-K}{K-3}$

2. Calculate the slope of the perpendicular direction to $m_{RS}$:

$m_{\perp}=-\dfrac1{m_{RS}} = \dfrac{3-K}{5-K}$

3. Calculate the coordinates of the midpoint of RS:

$M\left(\dfrac{3+K}2\ ,\ \dfrac{k+5}2 \right)$

4. Calculate the equation through M perpendicular to RS:

$\dfrac{y-\frac{k+5}2}{x-\frac{3+K}2}=\dfrac{3-K}{5-K}$ ...... Solve for y:

$y= \dfrac{3-K}{5-K} \cdot x + \dfrac{8}{5-K}$

5. Plug in the coordinates of P into the last equation:

$2=0+\dfrac{8}{5-K}~\implies~10-2K=8~\implies~\boxed{K=1}$

3. Hello, nitinnitin2!

Another approach . . .

Find the value of $k$ if the point $P(0,2)$ lies on the perpendicular bisector
of the segment joining the points $Q(3,k)\text{ and }R(k,5).$

If $P$ lies on the perpendicular bisector of $QR$, then: . $PQ \:=\:PR$

. . $\begin{array}{cccccccc}PQ &=& \sqrt{(3-0)^2 + (k-2)^2} & \Rightarrow & PQ^2 &=& k^2-4k+13 & {\color{blue}[1]} \\ \\[-3mm]PR &=& \sqrt{(k-0)^2 + (5-2)^2} & \Rightarrow & PR^2 &=& k^2+9 & {\color{blue}[2]} \end{array}$

Equate [1] and [2]: . $k^2-4k+13 \:=\:k^2+9 \quad\Rightarrow\quad\boxed{ k \:=\:1}$