Question

name two planes that intersect with \overleftrightarrow{de}.
diagram of a rectangular prism with points a, b, c, d, e, f, g, h
solve for x
number line with a, b, c; ab=3x, bc=6, ac=9; equations: 3x + 6 = 9, 3x = 3, x=1
find the distance between the points:
(-7, -9) and (2, 9)

Explanation:

First Problem (Name two planes intersecting with $\overleftrightarrow{DE}$):

Step1: Analyze the rectangular prism

In the given rectangular prism (a box - like figure), the line $\overleftrightarrow{DE}$ is part of some faces (planes) of the prism.

Step2: Identify planes containing $\overleftrightarrow{DE}$

• Plane $DEFG$: This plane has the line segment $DE$ as one of its edges. So the line $\overleftrightarrow{DE}$ lies on plane $DEFG$.
• Plane $DEHA$: This plane also contains the line segment $DE$ as one of its edges. So the line $\overleftrightarrow{DE}$ lies on plane $DEHA$.

Step1: Set up the equation

From the segment $AC$ with $AB = 3x$ and $BC=6$ and $AC = 9$, by the segment addition postulate, we have $AB + BC=AC$. So the equation is $3x + 6=9$.

Step2: Solve for $x$

Subtract 6 from both sides of the equation: $3x+6 - 6=9 - 6$, which simplifies to $3x = 3$. Then divide both sides by 3: $\frac{3x}{3}=\frac{3}{3}$, so $x = 1$.

Step1: Recall the distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Identify the coordinates

Here, $x_1=-7,y_1 = - 9,x_2 = 2,y_2=9$.

Step3: Substitute into the formula

First, calculate $(x_2 - x_1)=(2-(-7))=2 + 7=9$ and $(y_2 - y_1)=(9-(-9))=9 + 9 = 18$.
Then, $d=\sqrt{(9)^2+(18)^2}=\sqrt{81 + 324}=\sqrt{405}=\sqrt{81\times5}=9\sqrt{5}\approx9\times2.236 = 20.124$ (or we can leave it as $9\sqrt{5}$).

Answer:

Plane $DEFG$ and Plane $DEHA$ (other valid pairs like Plane $DEFG$ and Plane $CDEH$ etc. can also be correct depending on the prism's labeling, but these are two common ones)

Second Problem (Solve for $x$ in the segment addition):