Question

1. a) verify that ab and mn are parallel. b) verify that mn is half the length of ab. a(-2,4) b(4,-1) c(-1,3)

Explanation:

Step1: Recall the mid - point formula

The mid - point formula 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 $A(-2,4)$ and $B(4,-1)$. First, find the mid - point $M$ of $AC$ and $N$ of $BC$.
Let's assume $M$ is the mid - point of $AC$ and $N$ is the mid - point of $BC$.
If $A(-2,4)$ and $C(-1,3)$, then $M=(\frac{-2+( - 1)}{2},\frac{4 + 3}{2})=(-\frac{3}{2},\frac{7}{2})$
If $B(4,-1)$ and $C(-1,3)$, then $N=(\frac{4+( - 1)}{2},\frac{-1 + 3}{2})=(\frac{3}{2},1)$

Step2: Recall the distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For the distance between $A(-2,4)$ and $B(4,-1)$:
$AB=\sqrt{(4+2)^2+(-1 - 4)^2}=\sqrt{6^2+( - 5)^2}=\sqrt{36 + 25}=\sqrt{61}$
For the distance between $M(-\frac{3}{2},\frac{7}{2})$ and $N(\frac{3}{2},1)$:
$MN=\sqrt{(\frac{3}{2}+\frac{3}{2})^2+(1-\frac{7}{2})^2}=\sqrt{3^2+(-\frac{5}{2})^2}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36 + 25}{4}}=\frac{\sqrt{61}}{2}$
So $MN=\frac{1}{2}AB$.

Step3: Recall the slope formula

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.
The slope of $AB$ is $m_{AB}=\frac{-1 - 4}{4+2}=\frac{-5}{6}$
The slope of $MN$ is $m_{MN}=\frac{1-\frac{7}{2}}{\frac{3}{2}+\frac{3}{2}}=\frac{-\frac{5}{2}}{3}=-\frac{5}{6}$
Since $m_{AB}=m_{MN}$, $AB$ and $MN$ are parallel.

Answer:

We have verified that $AB$ and $MN$ are parallel and $MN$ is half the length of $AB$ by using the mid - point formula, distance formula and slope formula as shown above.