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# Equations of a straight line

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Hiiii!!! Could someone please explain how this works? May god bless you May 11, 2019

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If the slope of line DE  =  the slope of line BC, then DE  is parallel to  BC.

D  is the midpoint of AB, so    D   =   $$(\frac{4+2}{2},\frac{6-2}{2})$$   =   (3, 2)

E  is the midpoint of  AC, so   E   =   $$(\frac{4-2}{2},\frac{6-4}{2})$$   =   (1, 1)

slope of  DE  =  $$\frac{y_2-y_1}{x_2-x_1}\,=\,\frac{2-1}{3-1}\,=\,\frac{1}{2}$$

slope of  BC  =  $$\frac{y_2-y_1}{x_2-x_1}\,=\,\frac{-4+2}{-2-2}\,=\,\frac{-2}{-4}\,=\,\frac{1}{2}$$

Since the slope of  DE  and the slope of  BC  both equal  $$\frac12$$ ,  DE  is parallel to  BC.

Here's a graph to check:  https://www.desmos.com/calculator/5k2mf7pbjj

May 11, 2019

#1
+3

If the slope of line DE  =  the slope of line BC, then DE  is parallel to  BC.

D  is the midpoint of AB, so    D   =   $$(\frac{4+2}{2},\frac{6-2}{2})$$   =   (3, 2)

E  is the midpoint of  AC, so   E   =   $$(\frac{4-2}{2},\frac{6-4}{2})$$   =   (1, 1)

slope of  DE  =  $$\frac{y_2-y_1}{x_2-x_1}\,=\,\frac{2-1}{3-1}\,=\,\frac{1}{2}$$

slope of  BC  =  $$\frac{y_2-y_1}{x_2-x_1}\,=\,\frac{-4+2}{-2-2}\,=\,\frac{-2}{-4}\,=\,\frac{1}{2}$$

Since the slope of  DE  and the slope of  BC  both equal  $$\frac12$$ ,  DE  is parallel to  BC.

Here's a graph to check:  https://www.desmos.com/calculator/5k2mf7pbjj

hectictar May 11, 2019