Question

This is a two part question and it would really help me if you could solve both! :)

Answer 1

`\displaystyle\\Answer:\ none \ of\ these\ (m=-(5)/(3) );\ isosceles,\ right`

Explanation:

1.

a) find the midpoint G of the side DE:

`x_D=-2\ \ \ \ x_E=3\ \ \ \ y_D=-2\ \ \ \ y_E=1`

`\displaystyle\\x_G=(x_D+x_E)/(2) \\\\x_G=(-2+3)/(2)\\\\x_G=(1)/(2)\\\\x_G=0.5`

`\displaystyle\\y_G=(y_D+y_E)/(2)\\\\y_G=(-2+1)/(2) \\\\y_G=(-1)/(2) \\\\y_G=-0.5\\\\Thus,\ G(0.5,-0.5)`

b) find the midpoint I of the side DF:

`x_D=-2\ \ \ \ x_F=6\ \ \ \ y_D=-2\ \ \ \ y_F=-4`

`\displaystyle\\x_I=(x_D+x_F)/(2) \\\\x_I=(-2+6)/(2) \\\\x_I=(4)/(2) \\\\x_I=2`

`\displaystyle\\y_I=(y_D+y_F)/(2)\\\\y_I=(-2+(-4))/(2)\\\\y_I=(-6)/(2) \\\\y_I=-3\\\\Thus,\ I(2,-3)`

c) the slope of GI:

`x_G=0.5\ \ \ \ x_I=2\ \ \ \ y_G=-0.5\ \ \ \ y_I=-3`

`\displaystyle\\m_(GI)=(y_I-y_G)/(x_I-x_G) \\\\m_(GI)=(-3-(-0.5))/(2-0.5) \\\\m_(GI)=(-3+0.5)/(1.5) \\\\m_(GI)=(-2.5)/(1.5) \\\\m_(GI)=(-2.5(2))/(1.5(2)) \\\\m_(GI)=-(5)/(3)`

2.

Type of Δ DEF:

a) find the length of the side DE:

`|DE|=√((3-(-2)^2+(1-(-2)^2)\\\\|DE|=√((3+2)^2+(1+2)^2) \\\\|DE|=√(5^2+3^2)\\\\|DE|=√(25+9) \\\\|DE|=√(34) \ units`

b) find the length of the side EF:

`|EF|=√((6-3)^2+(-4-1)^2)\\\\|EF|=√(3^2+(-5)^2)\\\\ |EF|=√(9+25) \\\\|EF|=√(34)\ units`

Hence, DE=EF

c) find the m∠DEF:

`\displaystyle\\cos \angle E=\frac{\overrightarrow {DE}+\overrightarrow {EF}}DE \\\\`

Find the coordinates of the vector by the coordinates of its beginning and end points:

`\displaystyle\\\overrightarrow {DE}=(x_E-x_D,y_E-y_D)\\\\\overrightarrow {DE}=(3-(-2),1-(-2))\\\\\overrightarrow {DE}=(5,3)\\\\\overrightarrow {EF}=(x_F-x_E,y_F-y_E)\\\\\overrightarrow {EF}=(6-3),-5-1)\\\\\overrightarrow {EF}=(3,-5)\\Hence,\\\\cos\angle E=(5*3+3*(-5))/(√(34)*√(34) ) \\\\cos\angle E=(15-15)/(34 )\\\\cos\angle E=(0)/(34 )\\\\cos\angle E=0\\\\m\angle E=90^0`