Question

multiple choice
for exercises 1 - 5, choose the correct letter.

1. let (f(t)=2.5t + 10) represent the distance (f(t)) as a function of time (t) of a runner with a head - start in a race. another runner runs at the same rate but starts at the starting line. if (g(t)) represents the distance of the second runner as a function of time (t), what is true about the graphs of (f(t)) and (g(t))?

a. (f(t)) is perpendicular to (g(t)).
b. (f(t)) is parallel to (g(t)).
c. (f(t)) and (g(t)) are the same line.
d. (f(t)) intersects (g(t)).

1. which equation has a graph perpendicular to the graph of (7x-14y = 8)?

f. (y=-2x - 7) g. (y =-\frac{1}{2}x + 4) h. (y=\frac{1}{2}x - 1) j. (y = 2x+9)

1. which equation is the equation of a line that passes through ((-10,3)) and is parallel to (y = 5x-7)?

a. (y = 5x+53) b. (y = 5x - 47) c. (y=-\frac{1}{5}x + 1) d. (y=\frac{1}{5}x + 5)

1. which of the following coordinates for (p) will make (overline{mn}) perpendicular to (overline{qp}) in the diagram at the right?

f. ((-2,-5)) g. ((-3,0)) h. ((3,2)) j. ((3,5))

1. segment (xy) represents the path of an airplane that passes through the coordinates ((2,1)) and ((4,5)). what is the slope of a line that represents the path of another airplane that is traveling parallel to the first airplane?

a. (-2) b. (-\frac{1}{2}) c. (\frac{1}{2}) d. (2)
short response

1. a city designer is drawing the road map for a new housing development. palm st. runs through the coordinates ((11,5)) and ((-1,1)) on the map. pepperdine st. is going to run perpendicular to palm st. the coordinates of pepperdine st. are ((4,7)) and ((7,y)).

a. what is the value of (y)?

Explanation:

Step1: Recall slope - related rules

Parallel lines have equal slopes and perpendicular lines have slopes that are negative reciprocals of each other. The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Solve question 1

The functions $f(t)=2.5t + 10$ and $g(t)$ (same - rate runner) have the same slope. Since the slope of a linear function $y = mx + b$ is $m$, and same - slope lines are parallel, the answer is B.

Step3: Solve question 2

First, rewrite $7x-14y = 8$ in slope - intercept form $y=mx + b$. We get $y=\frac{1}{2}x-\frac{4}{7}$, so its slope is $\frac{1}{2}$. A line perpendicular to it has a slope of $- 2$. The equation $y=-2x - 7$ has a slope of $-2$, so the answer is F.

Step4: Solve question 3

The line $y = 5x-7$ has a slope of $5$. A line parallel to it has a slope of $5$. Using the point - slope form $y - y_1=m(x - x_1)$ with the point $(-10,3)$ and $m = 5$, we have $y-3=5(x + 10)$, which simplifies to $y=5x+53$. The answer is A.

Step5: Solve question 4

The slope of the line $MN$ with $M(-2,-5)$ and $N(4,7)$ is $m_{MN}=\frac{7+5}{4 + 2}=2$. A line perpendicular to it has a slope of $-\frac{1}{2}$. Let the slope of the line with endpoints $Q(-3,5)$ and $P(x,y)$. The slope formula gives $m_{QP}=\frac{y - 5}{x + 3}$. We check each option. For option H with $P(3,2)$, $m_{QP}=\frac{2 - 5}{3+3}=-\frac{1}{2}$, so the answer is H.

Step6: Solve question 5

The slope of the line passing through $(2,1)$ and $(4,5)$ is $m=\frac{5 - 1}{4 - 2}=2$. A line parallel to it has a slope of $2$, so the answer is D.

Step7: Solve question 6a

The slope of Palm St. passing through $(11,5)$ and $(-1,1)$ is $m_{Palm}=\frac{5 - 1}{11+1}=\frac{1}{3}$. The slope of Pepperdine St. (perpendicular to Palm St.) is $-3$. Using the slope formula for Pepperdine St. with points $(4,7)$ and $(7,y)$, we have $-3=\frac{y - 7}{7 - 4}$. Cross - multiply: $-3\times3=y - 7$, so $y=-2$.

Answer:

1. B. $f(t)$ is parallel to $g(t)$.
2. F. $y=-2x - 7$
3. A. $y = 5x+53$
4. H. $(3,2)$
5. D. $2$
6. a. $y=-2$