Analytic Geometry Q5

**looi76** · January 30th 2009, 11:31 PM

Question:
Show the points $A(-1,0)$ , $B(5,2)$ , $C(8,7)$ & $D(2,5)$ are vertices of a parallelogram.

I don't what I'm suppose to find the prove that it's a parallelogram. Should I find the slope m of the lines? or should I find the distance between the points? and what does vertices mean?!

EDIT: This question is related to the lesson of Inclination and Slope of a line.

**red_dog** · January 30th 2009, 11:50 PM

Yes, you have to find the slope of the lines and see that

$m_{AB}=m_{CD}$ and $m_{BC}=m_{AD}$

**looi76** · January 31st 2009, 12:30 AM

$A(-1,0)$ , $B(5,2)$ , $C(8,7)$ & $D(2,5)$

Slope of line $AB$ :

$A(-1,0)$ & $B(5,2)$

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2-0}{5-(-1)} = \frac{1}{3}$

Slope of line $CD$ :

$C(8,7)$ & $D(2,5)$

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5-7}{2-8} = \frac{1}{3}$

Slope of line $BC$ :

$B(5,2)$ & $C(8,7)$

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7-2}{8-5} = \frac{5}{3}$

Slope of line $AD$ :

$A(-1,0)$ & $D(2,5)$

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5-0}{2-(-1)} = \frac{5}{3}$

So, this is the answer?

**red_dog** · January 31st 2009, 12:53 AM

If $m_{AB}=m_{CD}\Rightarrow AB\parallel CD$

and if $m_{BC}=m_{AD}\Rightarrow BC\parallel AD$

So ABCD is a parallelogram.

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