Question

1. higher order thinking δefg and δhij have the same perimeter and side lengths. the coordinates are e(6, 2), f(9, 2), g(8, 7), h(0, 0), and i(4, 3).

a. what are possible coordinates of point j?
b. explain why there can be different possibilities for the coordinates for point j.

Explanation:

Step1: Calculate perimeter of $\triangle EFG$

First, find side lengths using distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$:
$EF=\sqrt{(9-6)^2+(2-2)^2}=\sqrt{3^2+0}=3$
$FG=\sqrt{(8-9)^2+(7-2)^2}=\sqrt{(-1)^2+5^2}=\sqrt{1+25}=\sqrt{26}$
$EG=\sqrt{(8-6)^2+(7-2)^2}=\sqrt{2^2+5^2}=\sqrt{4+25}=\sqrt{29}$
Perimeter $P=3+\sqrt{26}+\sqrt{29}$

Step2: Calculate known side of $\triangle HIJ$

$HI=\sqrt{(0-0)^2+(0-3)^2}=\sqrt{0+(-3)^2}=3$

Step3: Set up length equations for $IJ$ and $HJ$

Let $J=(x,y)$. We know either:
Case1: $IJ=\sqrt{26}$, $HJ=\sqrt{29}$
$\sqrt{(x-0)^2+(y-3)^2}=\sqrt{26}$ → $x^2+(y-3)^2=26$
$\sqrt{(x-0)^2+(y-0)^2}=\sqrt{29}$ → $x^2+y^2=29$
Subtract first equation from second: $x^2+y^2 - [x^2+(y-3)^2]=29-26$
$y^2-(y^2-6y+9)=3$ → $6y-9=3$ → $6y=12$ → $y=2$
Substitute $y=2$ into $x^2+y^2=29$: $x^2+4=29$ → $x^2=25$ → $x=5$ or $x=-5$
So $J=(5,2)$ or $(-5,2)$

Case2: $IJ=\sqrt{29}$, $HJ=\sqrt{26}$
$\sqrt{(x-0)^2+(y-3)^2}=\sqrt{29}$ → $x^2+(y-3)^2=29$
$\sqrt{(x-0)^2+(y-0)^2}=\sqrt{26}$ → $x^2+y^2=26$
Subtract second equation from first: $x^2+(y-3)^2 - (x^2+y^2)=29-26$
$(y^2-6y+9)-y^2=3$ → $-6y+9=3$ → $-6y=-6$ → $y=1$
Substitute $y=1$ into $x^2+y^2=26$: $x^2+1=26$ → $x^2=25$ → $x=5$ or $x=-5$
So $J=(5,1)$ or $(-5,1)$

When matching the perimeters, the unknown sides of $\triangle HIJ$ can be assigned to either of the two non-equal side lengths of $\triangle EFG$. Additionally, the x-coordinate can be positive or negative because the distance formula uses squared differences, so points with $\pm x$ will have the same distance from $H$ and $I$. This creates multiple valid positions for $J$ that satisfy the perimeter condition.

Answer:

a. The possible coordinates of point $J$ are $(5, 2)$, $(-5, 2)$, $(5, 1)$, $(-5, 1)$
b.