Question

1. given the points l, m, n, and o located at l(0, - 2), m(5,8), n(-1,5), and o(7,3) do the following:

(a) plot and label all four points.
(b) draw $overline{lm}$ and $overline{no}$.
(c) find the intersection point of $overline{lm}$ and $overline{no}$. label it as point p on the diagram. state the coordinates of p below.
(d) $angle mpn$ and $angle lpo$ are what type of angle pair?
(e) what does (d) tell you about the measures of $angle mpn$ and $angle lpo$?

1. given $overline{ef}$ has endpoints at e(-3,7) and f(7,1) do the following:

(a) plot and label e and f and draw $overline{ef}$.
(b) point g is located at g(2,4). plot and label g.
(c) what special point on $overline{ef}$ is point g?
(d) how could you justify your answer from (c)?
reasoning

1. $overline{ab}$ passes through a(-2,3) and b(4,6).

(a) plot a and b and draw $overline{ab}$.
(b) point c, at c(4,2), does not lie on $overline{ab}$. plot and label c. according to the parallel - line postulate, how many lines can be drawn through c that are parallel to $overline{ab}$?
(c) draw all the lines that pass through c and are parallel to $overline{ab}$. (hint: think about the slope of $overline{ab}$.)

Answer:

3.

(a)

To plot the points:

• For point $L(0, - 2)$, start at the origin $(0,0)$ and move 2 units down along the $y$-axis and label it as $L$.
• For point $M(5,8)$, start at the origin, move 5 units to the right along the $x$-axis and 8 units up along the $y$-axis and label it as $M$.
• For point $N(-1,5)$, start at the origin, move 1 unit to the left along the $x$-axis and 5 units up along the $y$-axis and label it as $N$.
• For point $O(7,3)$, start at the origin, move 7 units to the right along the $x$-axis and 3 units up along the $y$-axis and label it as $O$.

(b)

Use a straight - edge to draw a line segment connecting points $L$ and $M$ (label it as $\overline{LM}$) and another line segment connecting points $N$ and $O$ (label it as $\overline{NO}$).

(c)

1. First, find the equation of the line $\overline{LM}$:

• The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For points $L(0,-2)$ and $M(5,8)$, $m_{LM}=\frac{8-( - 2)}{5 - 0}=\frac{10}{5}=2$.
• Using the point - slope form $y - y_1=m(x - x_1)$ with the point $L(0,-2)$, the equation of the line $\overline{LM}$ is $y-(-2)=2(x - 0)$, which simplifies to $y = 2x-2$.

1. Then, find the equation of the line $\overline{NO}$:

• For points $N(-1,5)$ and $O(7,3)$, $m_{NO}=\frac{3 - 5}{7-( - 1)}=\frac{-2}{8}=-\frac{1}{4}$.
• Using the point - slope form with the point $N(-1,5)$, $y - 5=-\frac{1}{4}(x + 1)$, which expands to $y-5=-\frac{1}{4}x-\frac{1}{4}$, and then $y=-\frac{1}{4}x+\frac{19}{4}$.

1. Now, find the intersection point:

• Set the two equations equal to each other: $2x-2=-\frac{1}{4}x+\frac{19}{4}$.
• Multiply through by 4 to clear the fraction: $8x-8=-x + 19$.
• Add $x$ to both sides: $9x-8=19$.
• Add 8 to both sides: $9x=27$.
• Divide by 9: $x = 3$.
• Substitute $x = 3$ into $y = 2x-2$: $y=2\times3-2=4$. So the coordinates of point $P$ are $(3,4)$.

(d)

$\angle MPN$ and $\angle LPO$ are vertical angles. Vertical angles are formed when two lines intersect.

(e)

Vertical angles are congruent. So $m\angle MPN=m\angle LPO$.

4.

(a)

For point $E(-3,7)$, start at the origin, move 3 units to the left along the $x$-axis and 7 units up along the $y$-axis and label it as $E$. For point $F(7,1)$, start at the origin, move 7 units to the right along the $x$-axis and 1 unit up along the $y$-axis and label it as $F$. Then draw the line segment $\overline{EF}$.

(b)

For point $G(2,4)$, start at the origin, move 2 units to the right along the $x$-axis and 4 units up along the $y$-axis and label it as $G$.

(c)

Point $G$ is the mid - point of $\overline{EF}$.

(d)

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $E(-3,7)$ and $F(7,1)$, $\frac{-3 + 7}{2}=\frac{4}{2}=2$ and $\frac{7+1}{2}=\frac{8}{2}=4$. So the mid - point of $\overline{EF}$ is $(2,4)$, which is the location of point $G$.

5.

(a)

For point $A(-2,3)$, start at the origin, move 2 units to the left along the $x$-axis and 3 units up along the $y$-axis and label it as $A$. For point $B(4,6)$, start at the origin, move 4 units to the right along the $x$-axis and 6 units up along the $y$-axis and label it as $B$. Then draw the line segment $\overline{AB}$.

(b)

For point $C(4,2)$, start at the origin, move 4 units to the right along the $x$-axis and 2 units up along the $y$-axis and label it as $C$. According to the Parallel Line Postulate, exactly one line can be drawn through point $C$ that is parallel to $\overline{AB}$.

(c)

1. First, find the slope of $\overline{AB}$:

• Using the slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}$ for $A(-2,3)$ and $B(4,6)$, $m_{AB}=\frac{6 - 3}{4-( - 2)}=\frac{3}{6}=\frac{1}{2}$.

1. A line parallel to $\overline{AB}$ through point $C(4,2)$ will have the same slope $m=\frac{1}{2}$.

• Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(4,2)$ and $m=\frac{1}{2}$, we get $y - 2=\frac{1}{2}(x - 4)$, which simplifies to $y=\frac{1}{2}x$. Then draw this line passing through point $C$.