Which polygon does not belong with the others?

Mathematics

1 answer:

kvv77 [185]2 years ago

8 0

The second one. The octagon.

This is because all of the other choices are equal sided, which makes them a regular polygon.

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The verticle of a triangle are (2, 18) (-2, -4), and (6,12.

Sergeu [11.5K]

so the triangle has the vertices of (2, 18) (-2, -4), and (6,12), that gives us the endpoints for each line of

(2, 18) , (-2, -4)

(-2, -4) , (6,12)

(6,12) , (2, 18)

$\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{18})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-4}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-4-18}{-2-2}\implies \cfrac{-22}{-4}\implies \cfrac{11}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-18=\cfrac{11}{2}(x-2) \\\\\\ y-18=\cfrac{11}{2}x-11 \implies \blacktriangleright y=\cfrac{11}{2}x+7 \blacktriangleleft$

$\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-(-4)}{6-(-2)}\implies \cfrac{12+4}{6+2}\implies \cfrac{16}{8}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-4)=2[x-(-2)] \\\\\\ y+4=2(x+2)\implies y+4=2x+4\implies \blacktriangleright y=2x \blacktriangleleft$

$\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{12})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{18}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{18-12}{2-6}\implies \cfrac{6}{-3}\implies -2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-12=-2(x-6) \\\\\\ y-12=-2x+12\implies \blacktriangleright y=-2x+24 \blacktriangleleft$