Question

BC has one endpointB(3,2) and a midpointT(6,-2). Find thecoordinates of theother endpoint, C. Thank you in advance!!

Answer 1

We are given that BC has one endpoint B(3, 2) and a midpoint T(6, -2)

We are asked to find the coordinates of the other endpoint C.

Recall that the midpoint formula is given by

$T(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Where

$\begin{gathered} T(x,y)=(6,-2) \\ B(x_1,y_1)=(3,2) \end{gathered}$

So, the other endpoint C is

$x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

Substitute the given values,

$6=\frac{3+x_2}{2},-2=\frac{2+y_2}{2}$

Simplify,

$\begin{gathered} 2\cdot6=3+x_2,-2\cdot2=2+y_2 \\ 12=3+x_2,-4=2+y_2 \\ x_2=12-3,y_2=-4-2 \\ x_2=9,y_2=-6 \end{gathered}$

Therefore, the coordinates of the endpoint C are

$C(x_2,y_2)=(9,-6)$

Bonus:

Let us verify whether we got the correct coordinates or not.

$\begin{gathered} T(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ T(x,y)=(\frac{3+9}{2},\frac{2-6}{2}) \\ T(x,y)=(\frac{12}{2},\frac{-4}{2}) \\ T(x,y)=(6,-2) \end{gathered}$

Hence, we got the same midpoint as given in the question.