Respuesta :
I'll show you how to do the first exercises of both types, the others are identical.
Find the midpoint: A=(-2,-3), B=(8,-7)
The coordinates of the midpoint are the average of the correspondent coordinates, so we have
[tex]M=\left(\dfrac{A_x+B_x}{2}, \dfrac{A_y+B_y}{2}\right)=\left(\dfrac{-2+8}{2}, \dfrac{-3+-7}{2}\right)=(3,-5)[/tex]
Find the endpoint
We simply have to reverse the previous logic: since
[tex]M_x = \dfrac{A_x+B_x}{2}[/tex]
we have
[tex]B_x = 2M_x-A_x[/tex]
and the same goes for [tex]B_y[/tex].
So, we have
[tex]B_x = 2\cdot 3+1=7[/tex]
[tex]B_y = 2\cdot 4-2=6[/tex]
Answer:
[tex]1) (3,-5)\,,\,2)\left ( \frac{3}{2},-5 \right )\,,\,3)\left ( 5,\frac{1}{2} \right )\,,\,4)\left ( \frac{-9}{2},2 \right )\,,\,5)\left ( 7,6 \right )\,,\,6)(4,-3)\,,\,7)(8,9)[/tex]
Step-by-step explanation:
For points [tex]A(x_1,x_2)\,,\,B(y_1,y_2)[/tex] , midpoint is given by [tex]M\left ( \frac{x_1+y_1}{2},\frac{x_2+y_2}{2} \right )[/tex]
1) [tex]A(-2,-3)\,,\,B(8,-7)[/tex]
[tex]M=\left ( \frac{-2+8}{2},\frac{-3-7}{2} \right )=\left ( 3,-5 \right )[/tex]
2) [tex]A(9,-1)\,,\,B(-6,-11)[/tex]
[tex]M=\left ( \frac{9-6}{2},\frac{-1-11}{2} \right )=\left ( \frac{3}{2},-5 \right )[/tex]
3) [tex]A(-4,1)\,,\,B(14,0)[/tex]
[tex]M=\left ( \frac{-4+14}{2},\frac{1+0}{2} \right )=\left ( 5,\frac{1}{2} \right )[/tex]
4) [tex]A(0,-5)\,,\,B(-9,9)[/tex]
[tex]M=\left ( \frac{0-9}{2},\frac{-5+9}{2} \right )=\left ( \frac{-9}{2},2 \right )[/tex]
5) [tex]M(3,4)\,,\,A(-1,2)[/tex]
[tex](3,4) =\frac{1}{2}\left [ \left ( -1,2 \right )+\left ( x,y \right ) \right ]\\ \left ( 6,8 \right ) =\left ( -1+x,2+y \right )\\ -1+x =6\,,\,2+y=8\\ [/tex]
[tex]x=7,y=6[/tex]
6) [tex]M(1,0)\,,\,A(-2,3)[/tex]
[tex]\displaystyle (1,0) =\frac{1}{2}\left [ \left (-2,3 \right )+\left ( x,y \right ) \right ]\\\displaystyle \left ( 2,0\right ) =\left ( -2+x,3+y \right )\\\displaystyle -2+x =2\,,\,3+y=0\\[/tex]
x=4 , y= - 3
7) [tex]M(6,5)\,,\,A(4,1)[/tex]
[tex]\displaystyle (6,5) =\frac{1}{2}\left [ \left (4,1 \right )+\left ( x,y \right ) \right ]\\\displaystyle \left ( 12,10\right ) =\left ( 4+x,1+y \right )\\\displaystyle 4+x =12\,,\,1+y=10\\\\[/tex]
x=8 , y=9