Question

Help me I don’t know how to solve these problems.
I need to show work.

Answer 1

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

`M=\left((A_x+B_x)/(2), (A_y+B_y)/(2)\right)=\left((-2+8)/(2), (-3+-7)/(2)\right)=(3,-5)`

Find the endpoint

We simply have to reverse the previous logic: since

`M_x = (A_x+B_x)/(2)`

we have

`B_x = 2M_x-A_x`

and the same goes for
`B_y`.

So, we have

`B_x = 2\cdot 3+1=7`

`B_y = 2\cdot 4-2=6`

Answer 2

`1) (3,-5)\,,\,2)\left ( (3)/(2),-5 \right )\,,\,3)\left ( 5,(1)/(2) \right )\,,\,4)\left ( (-9)/(2),2 \right )\,,\,5)\left ( 7,6 \right )\,,\,6)(4,-3)\,,\,7)(8,9)`

Explanation:

For points
`A(x_1,x_2)\,,\,B(y_1,y_2)` , midpoint is given by
`M\left ( (x_1+y_1)/(2),(x_2+y_2)/(2) \right )`

1)
`A(-2,-3)\,,\,B(8,-7)`

`M=\left ( (-2+8)/(2),(-3-7)/(2) \right )=\left ( 3,-5 \right )`

2)
`A(9,-1)\,,\,B(-6,-11)`

`M=\left ( (9-6)/(2),(-1-11)/(2) \right )=\left ( (3)/(2),-5 \right )`

3)
`A(-4,1)\,,\,B(14,0)`

`M=\left ( (-4+14)/(2),(1+0)/(2) \right )=\left ( 5,(1)/(2) \right )`

4)
`A(0,-5)\,,\,B(-9,9)`

`M=\left ( (0-9)/(2),(-5+9)/(2) \right )=\left ( (-9)/(2),2 \right )`

5)
`M(3,4)\,,\,A(-1,2)`

`(3,4) =(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\\`

`x=7,y=6`

6)
`M(1,0)\,,\,A(-2,3)`

`\displaystyle (1,0) =(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\\`

x=4 , y= - 3

7)
`M(6,5)\,,\,A(4,1)`

`\displaystyle (6,5) =(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\\\\`

x=8 , y=9