Using Your Mathematics Toolbox A Culminating Assignment for EQUATION OF LINES

Dr. Robert E. Mason IV

2

Introduction

In this assignment we take a systematic look at equation of lines and their graphs. We begin by recalling the concept of slope, which you’ve seen in previous work in this class. The slope of a no vertical line is a number that measures the slant or direction of the line; it is defined as follows, The slope of a no vertical line passing through the two points (x1, y1) and (x2, y2) is the

number m defined by

Note that the quantity appearing in the definition of slope is the amount by which x changes as we move from to along the line. We denote this change is x by the symbol (read delta x). Thus = . Similarly, the symbol is defined to mean the change in y: = . Using these ideas, we can rewrite our definition of slope as m = / . The slope of a line does not depend on which two particular points on the line are used in the calculation. To see why this is so, consider the follow picture.

The two triangles are similar (because the corresponding angles are equal). This

implies that the corresponding sides of the two triangles are proportional, and so we have

�

m = y2 − y1x2 − x1

�

x2 − x1

�

(x1 ,y1)

�

(x2 ,y2)

�

Δx

�

Δx

�

x2 − x1

�

Δy

�

Δy

�

y2 − y1

�

Δy

�

Δx

�

ab

= bc

3

Now notice that the left-hand side of this equation represent the slope / calculated using the points A and D, and the right-hand side represents the slope calculated using the points B and C. Thus the values we obtain for the slope are indeed equal. HORIZONTAL AND VERTICAL LINES

1. The slope of a horizontal line is zero. 2. Slope is not defined for vertical lines.

THE POINT-SLOPE FORMULA

The equation of a line that has slope m and passes through the point is

THE SLOPE-INTERCEPT FORMULA

The equation of a line with slope m and y-intercept b is PARALLEL AND PERPENDICULAR LINES Let and denote the slopes of two no vertical lines. Then:

1. The lines are parallel if and only if ; and 2. The lines are perpendicular if and only if .

In Problems 1-3, compute the slope of the line passing through the two given points. In Problem 3, include a sketch with your answers. Show all work.

1. (a) (-3, 2), (1, -6) (b) (2, -5) (4, 1) (c) (-2,7), (1, 0) (d) (4, 5), (5, 8)

2. a) (-3, 0), (4, 9)

(b) (-1, 2), (5, 8) (c) (1/2, -3/5), (3/2, 3/4) (d) (17/3, -1/2). (-1/2, 17/3)

3. (a) (1, 1), (-1, -1)

(b) (0,5), (-8, 5) (c) (-1, 1), (1, -1) (d) (a, b), (b, a) (Assume a b)

�

Δy

�

Δx

�

(x1 ,y1)

�

y − y1 = m(x − x1)

�

y = mx + b

�

m1

�

m2

�

m1 = m2

�

m1 = −1m2

�

≠

4

4. Compute the slope of the line in the following figure using each pair of points indicated - (a) A and B (b) B and C (c) A and C. The principle involved here is that no matter which pair of points you choose, the slope is the same.

5. The slopes of four lines are indicated in the figure below. List the slopes , , ,

in order of increasing value.

�

m1

�

m2

�

m3

�

m4

5

6. Refer to the accompanying figure.

(a) List the slopes , , and in order of increasing size. (b) List the numbers in order of increasing size.

In Problems 7-9, three points A, B, and C are specified. Determine if A, B, and C are collinear (lie on the same line) by checking to see whether the slope of equals the slope of .

7. A(-8, -2), B(2, 1/2), C(11, -1) 8. A(4, -3), B(-1, 0), C(-4,2) 9. A(0, -5), B(3, 4), C(-1, -8) 10. If the area of the “triangle” formed by three points is zero, then the points must

in fact be collinear. Use this observation, along with the formula

, to rework Problem 9.

In Problems 11 and 12, find the equation for the line having the given slope and passing through the given point. Write your answers in the form .

11. (a) m = -5; (-2, 1) (b) m =4; (4, -4) (c) m = 1/3; (-6, -2/3) (d) m = -1; (0,1)

12.(a) m = 22; (0,0) (b) m = -222; (0, 0) (c) m = ; (0, 0)

�

m1

�

m2

�

m3

�

b1,b2 ,and b3

�

AB

�

BC

�

A = 12

x1y2 − x2y1 + x2y3 − x3y2 + x3y1 − x1y3

�

y = mx + b

�

2

6

In Problems 13 and 14, find the equation for the line passing through the two points. Write your answer in the form of .

13. (a) (4, 8) and (-3, -6) (b) (-2, 0) and (3, -10) (c) (-3, -2) and (4, -1) 14. (a) (7, 9) and (-11, 9) (b) (5/4, 2) and (3/4, 3) (c) (12, 13) and (13, 12)

In Problems 15 and 16, write the equation of a vertical line passing through the given point. In Problems 17 and 18, write the equation of a horizontal line passing through the given point.

15. (-3, 4) 16. (8, 5) 17. (-3, 4) 18. (8,5) 19. Is the graph of the line x = 0 the x-axis or the y-axis? 20. Is the graph of the line y = 0 the x-axis or the y-axis?

�

y = mx + b

7

In Problems 21-22, find the equation of the line with the given slope and y-intercept.

21. (a) slope –4; y-intercept 7 (b) slope 2; y-intercept 3/2 (c) slope –4/3; y-intercept 14

22. (a) slope 0; y-intercept 14 (b) slope 14; y-intercept 0

In Problems 23-28, find an equation for the line that is described, and sketch the graph. For Problems 23-24, write the final answer in the form ; for Problems 25-26, write the answer in the form Ax +By + C = 0.

23. (a) Passes through (-3, -1) and has slope 4 (b) Passes through (5/2, 0) and has slope 1/2 (c Has x-intercept 6, y-intercept 5 (d) Has x-intercept –2, and slope 3/4 (e) Passes through (1, 2) and (2,6)

24. (a) Passes through (-7, -2) and (0, 0) (b) Passes through (6, -3) and has y-intercept 8 (c) Passes through (0, -1) and has the same slope as the line 3x + 4y = 12

(d) Passes through (6, 2) and has the same x-intercept as the –2x + y = 1 (e) Has x-intercept –6, y-intercept 25. Passes through (-3, 4) and is parallel to the x-axis 26. Passes through (-3, 4) and is parallel to the y-axis

In Problems 27 and 28, find the x- and y-intercepts of the line, and find the area and perimeter of the triangle formed by the given line and the axes.

27. (a) 3x + 5y = 15 (b) 3x – 5y = 15 28. (a) 5x + 4y = 40 (b) 2x + 4y =

29. Determine whether each pair of lines is parallel, perpendicular, or neither.

a. 3x – 4y =12; 4x – 3y = 12 b. y = 5x – 16; y = 5x + 2 c. 5x – 6y = 25; 6x + 5y = 0 d. y = -2/3x –1; y = 3/2x – 1 e. –2x – 5y = 1; y – 2/5x –4 = 0 f. x = 8y + 3; 4y – 1/2 x = 32

30. Are the lines y = x + 1 and y = 1 – x parallel, perpendicular, or neither?

In Problems 31-35, find an equation for the line that is described. Write the answer in the forms y = mx + b and Ax + By + C = 0.

31. Is parallel to 2x – 5y = 10 and passes through (-1, 2) 32. Is parallel to 4x + 5y = 20 and passes through (0, 0)

�

y = mx + b

�

2

�

2

8

33. Is perpendicular to 4y – 3x = 1 and passes through (4, 0) 34. Is perpendicular to x – y + 2 = 0 and passes through (3, 1) 35. Is parallel to 3x – 5y = 25 and has the same y-intercept as the line 6x – y +

11 = 0

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Mind Bender 36. Follow my instructions in parts (a)-(e) to determine the slope m of the line in the figure. THE AREA OF THE TRIANGLE IS 4 SQ. UNITS.

(a) Use the point –slope formula to show that the equation of the line y –1 = m(x –2).

(b) Show that the y-intercept for the line is –2m + 1.

(c) Show that the area of the triangle, which is one-half base time height, is

(d) Set the expression for area in (c) equal to 4, as given in the problem, and simplify the resulting equation to obtain

(e) Solve the equation for m. You

should obtain m = -1/2. 37. A line with a slope of –5 passes through the point (3, 6). Find the area of the

triangle in the first quadrant formed by this line and the coordinate axes. Hint: First find the equation of the line.

38. The y-intercept of the line in the figure is 6. Find the slope of the line if the area of the shaded triangle is 72 square units.

�

12

(2m−1m

)(−2m +1).

�

4m2 + 4m +1= 0.

10

39. (a) Sketch the line and the point P(1, 3). Follow parts (b)-(d) to

calculate the perpendicular distance from point P(1, 3) to the line. (b) Find the equation of the line that passes through P(1, 3) to the line. (c) Find the coordinates of the point where these two lines intersect. (d) Use the distance formula to find the perpendicular distance from P(1, 3) to the

.

DISTANCE FROM A POINT TO A LINE

We have developed and use the formula for the distance between two points. The distance formula: . Now we consider another kind of distance formula, one that gives the distance d from a point to a line. As indicated in the figure below, distance in this context means the shortest distance, which is the perpendicular distance. In the box that follows, we show two equivalent forms for this formula. Although the second form is more widely known, the first is just as useful and somewhat simpler to derive. (We will not derive either).

a. The distance d from the point to the line is given by

b. The distance d from the to the line is given by

Example

Find the distance from the point (-3, 1) to the line y = -2x +7. To find the distance from the point (-3, 1) to the line y = -2x +7, we use the form one with = -3, = 1, m = -2, and b = 7. This yields

�

y = 12

x − 5

�

y = 12

x − 5

�

d = (x2 − x1)2 + (y2 − y1)2

�

(x0 ,y0)

�

y = mx + b

�

d =mx0 + b− y01+ m2

�

(x0 ,y0)

�

Ax + By + C = 0

�

d =Ax0 + By0 + C

A2 + B2

�

x0

�

y0

�

d =(−2)(−3) + 7 −1

1+ (−2)2= 12

5= 12 5

5units

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Exercises 40-44 are practice problem. To solve these problems, you will need to utilize the formulas listed at the beginning of this assignment. 40 Find the distance between the points (-5, -6) and (3, -1). 41. Find the equation of a line that is perpendicular to the line 4x-5y-20=0 and has the same y-intercept as the line x – y + 1 = 0. Write your answer in the form Ax + By + C = 0.

42. Find the equation of the line that passes through (2, -4) and is parallel to the line 3x – y = 1. Write your answer in the form y = mx + b. 43. Find the equation of the line that is the perpendicular bisector of the line segment joining the points (2, 1) and (6, 7). Write your answer in the form Ax + By + C = 0.

44. Find the perimeter of in the following figure.

In Exercises 45-50, find the distance from the point to the line using (a) the

; (b) the formula .

45. (1, 4); y = x – 2 46. (-2, -3); y = -4x + 1 47. (-3, 5); 4x + 5y + 6 = 0 48. (0, -3); 3x – 2y = 1 49. Find the distance between the two parallel lines y = 2x – 1 and y = 2x + 4. Hint: Draw a sketch; then find the distance from the origin to each line. 50. Find the distance between the two parallel lines 3x + 4y = 12 and 3x + 4y = 24.

�

ΔABC

�

d =mx0 + b− y01+ m2

�

d =Ax0 + By0 + C

A2 + B2