MAT039 Lesson 2: Lines and Circles PDF

MAT039 Lesson 2: LINES AND CIRCLES 2A: LINES MA. CRISTINA DUYAGUIT Department of Mathematics and Statistics, CSM Lines Definition 2.1: The (undirected) distance between any two points in the plane is defined as the length of the line segment joinin...

MAT039 Lesson 2: LINES AND CIRCLES 2A: LINES MA. CRISTINA DUYAGUIT Department of Mathematics and Statistics, CSM Lines Definition 2.1: The (undirected) distance between any two points in the plane is defined as the length of the line segment joining them. Theorem 2.1: (Distance Formula) Let 𝑃(𝑥1 , 𝑦1 ) and 𝑄(𝑥2 , 𝑦2 ) be two points in the real plane ℝ2. If 𝑑 is the distance between these point, then 𝑑 = 𝑃𝑄 = 𝑥2 − 𝑥1 2 + 𝑦2 − 𝑦1 2. Theorem 2.2: Let 𝑃(𝑥1 , 𝑦1 ) and 𝑄(𝑥2 , 𝑦2 ) be two points in the real plane ℝ2. If 𝑀 𝑥,ҧ 𝑦ത is the midpoint of the line segment joining the points 𝑃 and 𝑄, then 𝑥1 +𝑥2 𝑦1 +𝑦2 𝑥ҧ = and 𝑦ത = 2 2 MCLD, Department of Mathematics and Statistics 1 Lines Let 𝐿 be a non-horizontal line. Then 𝐿 intersects the 𝑥-axis. Denote by 𝛼 the angle formed by the line 𝐿 with the positive 𝑥-axis. The angle 𝛼 is called the angle of inclination of 𝐿 and it has a value between 0 and 𝜋 radians. 𝜋 Remark 2.1: If 𝐿 is vertical, then its angle of inclination is radians. We 2 define the angle of inclination of any horizontal line to be equal to zero radian. MCLD, Department of Mathematics and Statistics 2 Lines Definition 2.2: Let 𝐿 be a non-vertical line and let 𝛼 be its angle of inclination. Then the slope of 𝑳, denoted by 𝑚𝐿 , is given by 𝑚𝐿 = tan 𝛼. Theorem 2.3: Let 𝑃(𝑥1 , 𝑦1 ) and 𝑄(𝑥2 , 𝑦2 ) be two points in the plane with 𝑥1 ≠ 𝑥2. Then the slope of the line 𝐿 through the points 𝑃 and 𝑄 is given by 𝑦2 −𝑦1 𝑚𝐿 =. 𝑥2 −𝑥1 MCLD, Department of Mathematics and Statistics 3 Lines Examples 2.1: 1. Find the distance between the points (3, −2) and (0, 2). Solution: Let (3, −2) be the point 𝑃(𝑥1 , 𝑦1 ) and (0, 2) be the point Q(𝑥2 , 𝑦2 ). Using the distance formula, 𝑑= 0−3 2 + 2 − −2 2 = 9 + 16 = 5 units MCLD, Department of Mathematics and Statistics 4 Lines 2. The midpoint of a line segment 𝑃𝑄 is at (−2, −1). If the point 𝑃 has coordinates (5, −4) , find the coordinate of point 𝑄. Solution: Let (𝑥2 , 𝑦2 ) be the coordinates of 𝑄. Then 5+𝑥2 −4+𝑦2 −2 = and −1 =. 2 2 Solving for 𝑥2 and 𝑦2 , we get 𝑥2 = −9 and 𝑦2 = 2. Therefore, 𝑄 is the point −9, 2. MCLD, Department of Mathematics and Statistics 5 Lines 3𝜋 3. If a line 𝐿 has angle of inclination equal to radians, find its slope. 4 Solution: 3𝜋 Using the definition, 𝑚𝐿 = tan = −1. 4 4. Find the slope of the line 𝐿 through (−3, 4) and (2, −6). Solution: Let 𝑥1 , 𝑦1 = (−3, 4) and 𝑥2 , 𝑦2 = (2, −6). Then −6 − 4 10 𝑚𝐿 = =− = −2 2 − (−3) 5 MCLD, Department of Mathematics and Statistics 6 Lines Equation of a Line 𝑃(𝑥1 , 𝑦1 ) and 𝑄(𝑥2 , 𝑦2 ) – two points on line 𝐿, 𝑚 is the slope of line 𝐿, 𝑎 is the 𝑥-intercept of 𝐿, 𝑏 is 𝑦-intercept of 𝐿 1. point-slope form (given a point on the line and the slope) 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 2. slope-intercept form (given the slope and the y-intercept of the line) 𝑦 = 𝑚𝑥 + 𝑏 3. two-point form (given two points of the line) 𝑦2 −𝑦1 𝑦 − 𝑦1 = (𝑥 − 𝑥1 ) 𝑥2 −𝑥1 4. Intercepts form (given the 𝑥 and 𝑦-intercept of the line) 𝑥 𝑦 + =1 𝑎 𝑏 Department of Mathematics and Statistics 7 Lines Examples 2.2: 1. Find the slope and the 𝑦-intercept of the line 7𝑥 + 3𝑦 − 6 = 0. Solution: Transform the equation of the line 7𝑥 + 3𝑦 − 6 = 0 (which is in general form) in the slope-intercept form, we get 7 𝑦 =− 𝑥+2. 3 7 Therefore, the slope is 𝑚 = − and the 𝑦-intercept is (0, 2). 3 MCLD, Department of Mathematics and Statistics 8 Lines 2. Find an equation f the line through the origin and the midpoint of the line segment joining (2, 3) and (4, 1). Solution: First we find the midpoint 𝑀 𝑥,ҧ 𝑦ത of the line segment joining (2, 3) and 2+4 3+1 (4, 1) : 𝑥ҧ = = 3 and 𝑦ത = = 2. 2 2 Next we find the slope of the line through 𝑥1 , 𝑦1 = 0, 0 and 2−0 2 𝑥2 , 𝑦2 = 𝑀 𝑥,ҧ 𝑦ത = 3, 2 : 𝑚 = 3−0 = 3. Using the point-slope form of an equation of a line and using the point of origin (0, 0), the equation of the line is: 2 𝑦 − 0 = 3 (𝑥 − 0), 2 which is equivalent to 𝑦 = 𝑥 or 2𝑥 − 3𝑦 = 0. 3 MCLD, Department of Mathematics and Statistics 9 Lines 3. The 𝑦-intercept of line 𝐿 is (0, −3). If 𝐿 passes through the intersection of the lines 2𝑥 + 3𝑦 + 1 = 0 and 𝑥 + 2𝑦 − 1 = 0, find an equation of 𝐿. Solution: First we find the point of intersection of the two given lines, i.e., 2𝑥 + 3𝑦 + 1 = 0 (1) we solve the system of equations ቊ : 𝑥 + 2𝑦 − 1 = 0 (2) multiply equation (2) with 2 and subtract the new equation from equation (1), we have MCLD, Department of Mathematics and Statistics 10 Lines Cont’n of solution of Examples 2.2 #3 2𝑥 + 3𝑦 + 1 = 0 − (2𝑥 + 4𝑦 − 2 = 0) ____________________ −𝑦 + 3 = 0 or 𝑦 = 3. Substitute 𝑦 = 3 to equation (1), we get 𝑥 = −5. Hence the intersection point of the two given lines is (−5, 3). Let 𝑥1 , 𝑦1 = 0, −3 and 𝑥2 , 𝑦2 = (−5, 3). Using the two-point form of an equation of a line, the equation of line 𝐿 is: 3− −3 𝑦 − −3 = (𝑥 − 0), equivalent to 6𝑥 + 5𝑦 + 15 = 0. −5−0 MCLD, Department of Mathematics and Statistics 11 Lines Theorem 2.4: Let 𝐿 and 𝑙 be two lines which are not vertical. Then 𝐿 and 𝑙 are parallel if and only if 𝑚𝐿 = 𝑚𝑙. Theorem 2.5: Let 𝐿 and 𝑙 be two lines which are not vertical. Then 𝐿 and 𝑙 1 are perpendicular if and only if 𝑚𝐿 = −. 𝑚𝑙 Definition 2.3: Let 𝐿 be a line and 𝑃 be a point not on the line 𝐿. The distance from 𝑷 to the line 𝑳 is the length of the perpendicular line segment ( shortest line segment) dropped from point 𝑃 to the line 𝐿. Theorem 2.6: Let 𝐿 be the line 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 and 𝑃(𝑥0 , 𝑦0 ) be a point in the plane. The distance 𝑑 from the point 𝑃 to the line 𝐿 is given by 𝐴𝑥0 + 𝐵𝑦0 + 𝐶 𝑑 = 𝑑 𝑃, 𝐿 = 𝐴2 + 𝐵2 MCLD, Department of Mathematics and Statistics 12 Lines Examples 2.3: 1. Find an equation of the line passing through the point (−4, −1) and parallel to the line 3𝑥 − 7𝑦 + 2 = 0. Solution: Let 𝑃 𝑥1 , 𝑦1 = (−4, −1) , 𝑙: 3𝑥 − 7𝑦 + 2 = 0 , and 𝐿 be the unknown line. 3 2 Solving for 𝑦 in the equation of 𝑙: 𝑦 = 𝑥 +. 7 7 3 3 Therefore, 𝑚𝑙 =. Since 𝑙 and 𝐿 are parallel, 𝑚𝐿 = 𝑚𝑙 =. 7 7 Using the point-slope form of an equation for a line, the equation of line L 3 is : 𝑦 − −1 = (x − −4 ), equivalent to 3𝑥 − 7𝑦 + 5 = 0. 7 MCLD, Department of Mathematics and Statistics 13 Lines 2. Find an equation of the line through the (5, −3) and perpendicular to the line through the points (−3, −1) and (1, 5). Solution: Let 𝐿 be the unknown line and 𝑙 be the line passing through the 5−(−1) 6 3 points (−3, −1) and (1, 5). Then 𝑚𝑙 = = =. Since 𝐿 is 1−(−3) 4 2 1 2 perpendicular to 𝑙, we have 𝑚𝐿 = − =−. 𝑚𝑙 3 Let 𝑃 𝑥1 , 𝑦1 = (5, −3), then using the point-slope form of an equation of a 2 line, the equation of line 𝐿 is : 𝑦 − −3 = − 𝑥 − 5 , or equivalently, 3 2𝑥 + 3𝑦 − 1 = 0. MCLD, Department of Mathematics and Statistics 14 Lines 3. Find the two points on the 𝑥-axis which are at a unit distance from the line 3𝑥 + 4𝑦 + 5 = 0. Solution: Let 𝑥0 , 0 be a point on the 𝑥-axis which is at a distance 𝑑 = 1 unit from the line 3𝑥 + 4𝑦 + 5 = 0. From the distance formula, we 3 𝑥0 +4 0 +5 3𝑥0 +5 3𝑥0 +5 have: 1= = =. 32 +4 2 25 5 That is, 5 = 3𝑥0 + 5. This implies that 3𝑥0 + 5 = 5 or 3𝑥0 + 5 = −5. 10 Solving for 𝑥0 : 𝑥0 = 0 or 𝑥0 = −. Therefore, the two points are 0, 0 3 10 and − ,0. 3 MCLD, Department of Mathematics and Statistics 15 Lines Exercises 2.1: A. Solve as indicated: 1. The midpoint of the line segment 𝑃𝑄 is at the point (3, 4). If the abscissa of 𝑃 is 4 and the ordinate of 𝑄 is −2 , find the points 𝑃 and 𝑄. 2. Show that the quadrilateral with vertices 𝐴(1, 4), 𝐵(1, −3), 𝐶(−6, 4) and 𝐷(−6, −3) is a square. 3. Find the distance form (1, 2) to the midpoint of the line segment whose endpoints are (−1, −3) and (−3, −5). 4. Find 𝑎 such that the distance from (−1, 2) to 𝑎𝑥 − 𝑦 + 9 = 0 is 5 units. 5. Find the shortest distance between the two parallel line with equations 2𝑥 − 𝑦 − 2 = 0 and 2𝑥 − 𝑦 − 7 = 0. MCLD, Department of Mathematics and Statistics 16 Lines Exercises 2.1: B. Find the slope and 𝑦-intercept of the following lines: 1. 𝑦=6 2. 𝑥=0 3. −7𝑥 = −6𝑦 + 5 4. 2𝑥 + 5𝑦 + 1 = 0 1 5. 𝑥 + 5𝑦 = 7 2 MCLD, Department of Mathematics and Statistics 17 Lines Exercises 2.1: C. Find an equation of the line satisfying the given conditions. 5𝜋 1. angle of inclination is radians and the ordinate of the 𝑦-intercept is 6 −4 3 2. slope is − and passes through the midpoint of the line segment 4 connecting (−4, 2) and (3, −3) 3. passes through (−2, −2) and parallel to the 𝑥-axis 4. perpendicular to 2𝑥 + 3𝑦 − 5 = 0 and the 𝑥-intercept is (−2, 0) 5. perpendicular to 𝐿: 𝑥 + 𝑦 + 2 = 0 and passing through the midpoint of the line segment joining the 𝑥 and 𝑦-intercepts of line 𝐿 MCLD, Department of Mathematics and Statistics 18 References: · Canoy, Sergio, Jr., et al. A First Course in Analytic Geometry and Calculus. Revised Edition. Department of Mathematics, MSU-IIT. 2010. · Leithold, I. (1996). The Calculus 7. HarpenCollins College Publishing. · Leithold, Louis. The Calculus with Analytic Geometry, 7th edition, 1995. · Mendelson, Elliott. 3,000 Solved Problems in Calculus. Mc-Graw Hill, 1988. · Protter, M &Protter, P. (1988). Calculus with Analytical Geometry. Boston: Jones and Bartlett Publisher. · Protter, Murray H. and Morrey, Charles B. Calculus with Analytic Geometry. Addison-Wesley Educational Publishers Inc., 1971. · Purcell, E. & Patterson, R. (1978). Calculus with Analytic Geometry. Prentice-Hall.