Answer to Question #277168 in Analytic Geometry for Jovy

Question

1. Solve for the area in square units of the right triangle with vertices F(0,7),U(2,-3),N(7,-2)

2. Find the area in square units of a rectangle whose vertices are L(0,6),O(2,-2),V(-2,-3),E(-4,5)

Answer 1

1.

$\overrightarrow{UF}=\langle0-2, 7-(-3)\rangle=\langle-2,10)\rangle$

$\overrightarrow{UN}=\langle7-2, -2-(-3)\rangle=\langle5,1)\rangle$

$\overrightarrow{UF}\cdot\overrightarrow{UN}=-2(5)+10(1)=0=>\overrightarrow{UF}\perp\overrightarrow{UN}$

The right triangle $FNU$

$|\overrightarrow{UF}|=\sqrt{(-2)^2+(10)^2}=2\sqrt{26}$

$|\overrightarrow{UN}|=\sqrt{(5)^2+(1)^2}=\sqrt{26}$

$S=\dfrac{1}{2}|\overrightarrow{UF}||\overrightarrow{UN}|=\dfrac{1}{2}(2\sqrt{26})(\sqrt{26})=26({units}^2)$

2.

$\overrightarrow{VO}=\langle2-(-2), -2-(-3)\rangle=\langle4,1)\rangle$

$\overrightarrow{VE}=\langle-4-(-2), 5-(-3)\rangle=\langle-2,8)\rangle$

$\overrightarrow{EL}=\langle0-(-4), 6-5\rangle=\langle4,1)\rangle=\overrightarrow{VO}$

$\overrightarrow{OL}=\langle0-2, 6-(-2)\rangle=\langle-2,8)\rangle=\overrightarrow{VE}$

$\overrightarrow{VE}\cdot\overrightarrow{VO}=-2(4)+8(1)=0=>\overrightarrow{VE}\perp\overrightarrow{VO}$

The rectangle $VELO$

$|\overrightarrow{VE}|=\sqrt{(-2)^2+(8)^2}=2\sqrt{17}$

$|\overrightarrow{VO}|=\sqrt{(4)^2+(1)^2}=\sqrt{17}$

$S=|\overrightarrow{VE}||\overrightarrow{VO}|=2\sqrt{17}(\sqrt{17})=34({units}^2)$