Rectangle EFGH has vertices E(1,-1), F(3,-5), G(7,-5), H(6,-1). If the shape is rotated 90 degrees clockwise about the origin, what are the coordinates of H'

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apowers6361

11.08.2021

Mathematics

check the picture below.

so then, its base is dc and we know its altitude or height is ef, so it has an area of (1/2)(dc)(ef).

$\bf \textit{distance between 2 points} d(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad c(\stackrel{x_2}{3}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} dc=\sqrt{)]^2+[-2-2]^2}\implies dc=\sqrt{(3+1)^2+(-4)^2} dc=\sqrt{16+16}\implies dc=\sqrt{32} [-0.35em] ~\dotfill$

$\bf \textit{distance between 2 points} e(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{1})\implies ef=\sqrt{(0-2)^2+(1-3)^2} ef=\sqrt{(-2)^2+(-2)^2}\implies ef=\sqrt{4+4}\implies ef=\sqrt{8} [-0.35em] \rule{34em}{0.25pt} \textit{\large area of the triangle}$

$\bf \cfrac{1}{2}(\sqrt{32})(\sqrt{8})\implies \cfrac{1}{2}(\sqrt{2^5})(\sqrt{2^3})\implies \cfrac{1}{2}\sqrt{2^5\cdot 2^3}\implies \cfrac{1}{2}\sqrt{2^8}\implies \cfrac{1}{2}\sqrt{(2^4)^2} \cfrac{1}{2}2^4\implies 2^3\implies 8$

Calculate the area of triangle cde with altitude ef, given c (3, −2), d (−1, 2), e (2, 3), and f (0,

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