Practice Problems

1. Estimate the slope of each line in Figure 19.

3. Calculate the slope of the line through each pair of points:

a) $( 2, 4 )$ , $( 5, 8 )$

b) $( –2, 4 )$ , $( 3, –5 )$

c) $( 2, 4 )$ , $( x, x2 )$

d) $( 2, 5 )$ , $( 2+h, 1+ (2+h)^2)$

e) $( x, x^2+ 3 )$ , $( a, a^2 + 3 )$


5. A small airplane at an altitude of 5000 feet is flying East at 300 feet per second (a bit over 200 miles per hour), and you are watching it with a small telescope as it passes directly overhead. (Fig. 21)

a) What is the slope of the telescope 5, 10 and 20 seconds after the plane passes overhead?

b) What is the slope of the telescope t seconds after the plane passes overhead?

c) After it passes overhead, is the slope of the telescope increasing, decreasing, or staying the same?


7. The blocks in a city are all perfect squares. A friend gives you the following directions to a good restaurant; "go north 3 blocks, turn east and go 5 blocks, turn south and go 7 blocks, turn west and go 3 blocks". How far away (straight line distance) is the restaurant?


9. How far up a wall will a 20 foot long ladder reach if the bottom must be at least 4 feet from the bottom of the wall? What will be the slope of the ladder if the bottom is 4 feet from the wall? What angle will the ladder make with the ground?

11. Let $P = (2 , 3)$ and $Q = (8 , 11)$. Verify that if $0 ≤ a ≤ 1$, then the point $R = (x,y)$ with $x = 2a + 8(1–a)$ and $y = 3a + 11(1–a)$ is on the line from $P$ to $Q$ and $Dist(P,R) = (1–a).Dist(P,Q)$.

13. The lines $y= x$ and $y= 4-x$ intersect at the point $(2, 2)$.

a) Use slopes to show that the lines are perpendicular.

b) Graph them together on your calculator using the "window" $-10 \leq x \leq 10$, $-10 \leq y \leq 10$. Why do the lines not appear to be perpendicular on the calculator display?

c) Find a suitable window for the graphs so the lines so that they do appear perpendicular.

15. Sketch each line which has $slope=m$ and which goes through the point $P$. Find the equation of each line.

a) $m = 3$, $P = (2,5)$

b) $m = –2$, $P = (3,2)$

c) $m = –1/2$, $P=(1,4)$

17. Find the equation of each of the following lines.

a) $L_1$ goes through the point $(2, 5)$ and is parallel to $3x – 2y = 9$.

b) $L_2$ goes through the point $(–1,2)$ and is perpendicular to $2x = 7–3y$.

c) $L_3$ goes through the point $(3, –2)$ and is perpendicular to $y = 1$.

19. Find the shortest distance between the circles with centers $C_1 = (1, 2)$ and $C_2 = (7, 10)$ and radii

a) $r_1 = 2$ and $r_2 = 4$

b) $r_1 = 2$ and $r_2 = 7$

c) $r_1 = 5$ and $r_2 = 8$

d) $r_1 = 3$ and $r_2 = 15$

e) $r_1 = 12$ and $r_2 = 1$

21. Explain how you can determine, without graphing, whether a given point $P = (x,y)$ is inside, on, or outside the circle with center $C = (h,k)$ and radius $r$.

23. Show that the equation of the circle with center $C = (h,k)$ and radius $r$ is $(x – h)^2 + (y – k)^2 = r^2$.

25. Find the slope of the line which is tangent to the circle with center $C = (3,1)$ at the point $P$ when

a) $P = (8,13)$

b) $P = (–10,1)$

c) $P = (–9,6)$

d) $P = (3,14)$

27.

a) How close does the line $3x – 2y = 4$ come to the point $(2,5$)?
b) How close does the line $y =5 – 2x$ come to the point $(1,–2)$?
c) How close does the circle with radius 3 and center at $(2,3)$ come to the point $(8,3)$?

29.

a) Show that the line L given by $Ax + By = C$ has $slope m = –A/B$. (Fig. 23)

b) Find the equation of the line $L*$ through $(0,0)$ which is perpendicular to line $L$ in part $(a)$.

c) Show that the lines $L$ and $L*$ intersect at the point

$(x, y) = ( \dfrac{AC}{A^2 + B^2} , \dfrac{BC}{A^2 + B^2})$.

d) Show that the distance from the origin to the point $(x,y)$ in part $(c)$ is

$\dfrac{| C |}{A^2 + B^2}$.

Steps (a) – (d) show that the distance from the origin to the line $Ax + By = C$ is $\dfrac{| C |}{A^2 + B^2}$.

1. (a) –3/4 (b) 1/2 (c) 0 (d) 2 (e) undefined

3. (a) $\dfrac{4}{3}$ (b) $\dfrac{ –9}{5}$ (c) $x + 2$ (if $x ≠ 2$) (d) $4 + h$ (if $h ≠ 0$) (e) $a + x$ (if $a ≠ x$)

5. (a) $t = 5: \dfrac{5000}{1500} = \dfrac{10}{3} , t = 10: \dfrac{5000}{3000} =\dfrac{5}{3} , t = 20: \dfrac{5000}{6000} =\dfrac{5}{6}$

(b) any $t > 0: \dfrac{5000}{300t} = \dfrac{50}{3t}$

(c) decreasing, since the numerator remains constant at 5000 while the denominator increases.

7. The restaurant is 4 blocks south and 2 blocks east. The distance is $\sqrt{4^2 + 2^2} =\sqrt{20} ≈ 4.47$ blocks.

9. $y = \sqrt{20^2 – 4^2} = \sqrt{384} ≈ 19.6$ feet, $m = \dfrac{\sqrt{384}}{4} ≈ 4.9 . tan( q ) = \dfrac{\sqrt{384}}{4} ≈ 4.9$ so $q ≈ 1.37 (≈ 78.5^o)$.

11. The equation of the line through $P = (2,3)$ and $Q = (8,11)$ is $y – 3 =\dfrac{8}{6} (x – 2)$ or $6y – 8x = 2$. Substituting $x = 2a + 8(1–a) = 8 – 6a$ and $y = 3a + 11(1–a) = 11 – 8a$ into the equation for the line, we get $6(11 – 8a) –8(8 – 6a) = 66 – 48a –64 + 48a$ which equals 2 for every value of a, so the point with $x = 2a + 8(1–a)$ and $y = 3a + 11(1–a)$ is on the line through P and Q for every value of a.

$The Dist(P,Q) =\sqrt{6^2 + 8^2} = 10$. $Dist(P,R) = \sqrt{(8–6a – 2)^2 + (11–8a – 3)^2}$

$=\sqrt{(6 – 6a)^2 + (8 – 8a)^2} = \sqrt{6^2(1–a)^2 + 8^2(1–a)^2} = \sqrt{100(1–a)^2} = 10 . |1–a| = |1–a| . Dist(P,Q)$.

13. (a) $m_1 \cdot m_2 = (1)(-1) = -1$ so the lines are perpendicular.

(b) Because 20 units of x-values are physically wider on the screen than 20 units of y-values.

(c) Set the window so (xmax - xmin) ≈ 1.7 (ymax - ymin).

15. (a) $y – 5 = 3(x – 2)$ or $y = 3x – 1$.

(b) $y – 2 = –2(x – 3)$ or $y = 8 – 2x$

(c) $y – 4 = –\dfrac{1}{2}(x – 1)$ or $y = –\dfrac{1}{2} x + \dfrac{9}{2}$

17. (a) $y – 5 = \dfrac{3}{2}(x–2)$ or $y = \dfrac{3}{2} x + 2$

(b) $y – 2 = \dfrac{3}{2}(x+1)$ or $y = \dfrac{3}{2} x + \dfrac{7}{2}$

(c) $x = 3$.

19. The distance between the centers is $\sqrt{6^2 + 8^2} = 10$.

(a) $10–2–4 = 4$

(b) $10–2–7 = 1$

(c) 0 (they intersect)

(d) $15–10–3 = 2$

(e) $12–10–1 = 1$.

21. Find $Dist( P,C ) = \sqrt{ (x–h)^2 + (y–k)^2}$, and compare the value to r: P is

inside the circle              if $Dist( P,C ) < r$

on the circle                   if $Dist( P,C ) = r$

outside the circle          if $Dist( P,C ) > r$


23. A point $P =(x,y)$ lies on the circle if and only if its distance from $C = (h,k)$ is $r : Dist( P,C ) = r$. So P is on the circle if and only if $\sqrt{(x–h)^2 + (y–k)^2} = r$ or $(x–h)^2 + (y–k)^2 = r^2$.

25. (a) $slope is –\dfrac{5}{12}$

(b) undefined (vertical line)

(c) $\dfrac{12}{5}$

(d) 0 (horizontal line)

27. (a) distance ≈ 2.22.

(b) Distance ≈ 2.24.

(c) (by inspection) 3 units which occurs at the point (5, 3).

29. (a) If $B ≠ 0$, we may solve for y: $y = – \dfrac{A}{B} x + \dfrac{C}{B}$. The slope is the coefficient of x: $m = –\dfrac{A}{B}$.

(b) The required slope is B/A (the negative reciprocal of –A/B) so the equation is $y = \dfrac{B}{A} x$ or $Bx–Ay = 0$.

(c) Solve ${ Ax + By = C, Bx – Ay = 0 }$ to get $x = \dfrac{AC}{A^2 + B^2}$ and $y = \dfrac{BC}{A^2 + B^2}$.

(d) Distance = $\sqrt { (\dfrac{AC}{(A^2 + B^2)})^2 + (\dfrac{BC}{A^2 + B^2})^2}$ = $\sqrt{\dfrac{A^2C^2}{(A^2 + B^2)^2} + \dfrac{B^2C^2}{(A^2 + B^2)^2}}$

$= \sqrt{\dfrac{(A^2 + B^2)C^2}{(A^2 + B^2)^2}} = \sqrt{\dfrac{C^2}{A^2 + B^2}} = \dfrac{| C |}{\sqrt{A^2 + B^2}}$