Question

1. find the distance between the coordinates (2, 3) and (-7, 0). -4, -3/2 9. the midpoint of line eg is at f(6, 8). if point e is located at (4, 0), where would point g be? your answer this is a required question 10. find the midpoint between the coordinates (-5, -3) and (-3, 0). your answer

Explanation:

Step1: Recall distance formula

The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here, $(x_1,y_1)=(2,3)$ and $(x_2,y_2)=(-7,0)$.

Step2: Substitute values

$d=\sqrt{(-7 - 2)^2+(0 - 3)^2}=\sqrt{(-9)^2+(-3)^2}=\sqrt{81 + 9}=\sqrt{90}=3\sqrt{10}$

Step3: Recall mid - point formula

The mid - point $M$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Let the coordinates of point $G$ be $(x,y)$. Given mid - point $F(6,8)$ and $E(4,0)$. Then $\frac{4 + x}{2}=6$ and $\frac{0 + y}{2}=8$.

Step4: Solve for $x$ and $y$ of point $G$

For $\frac{4 + x}{2}=6$, multiply both sides by 2: $4+x = 12$, so $x=8$. For $\frac{0 + y}{2}=8$, multiply both sides by 2: $y = 16$. So point $G$ is $(8,16)$.

Step5: Find mid - point of $(-5,-3)$ and $(-3,0)$

Using the mid - point formula $M=(\frac{-5+( - 3)}{2},\frac{-3 + 0}{2})=(\frac{-8}{2},\frac{-3}{2})=(-4,-\frac{3}{2})$

Answer:

1. $3\sqrt{10}$
2. $(8,16)$
3. $(-4,-\frac{3}{2})$