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Bài tập và giải pháp về Phương trình của đường thẳng và đường cong là tài liệu tập trung vào việc thực hành và nắm vững kiến thức về phương trình của đường thẳng và đường cong trong toán học. Tài liệu này cung cấp một loạt các bài tập về phương trình của đường thẳng, đường cong và các vấn đề liên quan. Mỗi bài tập đều được giải thích chi tiết và đi kèm với giải pháp minh họa để giúp bạn hiểu rõ cách giải quyết vấn đề.
Equations of Lines and Curves: Practice Problems and Solutions Question Proof The midpoint coordinates are ( ) given by the average of the corresponding coordinates of A and B 2+(−4) 3+1 Thus, the midpoint is , = (−1, 2) quod erat dem■ Question Determine the distance between the points P (3, −2) and Q(−1, 5) √ Proof The distance between two points P (x1 , y1 ) and Q(x2 , y2 ) is given by the formula (x2 − x1 )2 + (y2 − y1 )2 √ √ √ Therefore, the distance between P and Q is (−1 − 3)2 + (5 − (−2))2 = 64 + 49 = 113 quod erat dem■ Question Find the equation of the line passing through the points A(2, 4) and B(−1, −3) Proof The equation of a line passing through two points A(x1 , y1 ) and B(x2 , y2 ) is given by the formula y − y1 = xy22 −y (x − x1 ) Plugging in the values, we have y − = −3−4 (x − 2) Simplifying, we get −x1 −1−2 y = −x + quod erat dem■ Question Determine the slope of the line passing through the points P (2, 5) and Q(−3, 1) Proof The slope of a line passing through two points P (x1 , y1 ) and Q(x2 , y2 ) is given by the formula −y1 −4 1−5 m = xy22 −x = −5 = 45 Plugging in the values, we have m = −3−2 quod erat dem■ Question Find the equation of the line parallel to the x-axis and passing through the point P (3, −2) Proof Since the line is parallel to the x-axis, its equation is of the form y = k, where k is a constant Since the line passes through P (3, −2), we have −2 = k, so the equation is y = −2 quod erat dem■ Question Find the equation of the line perpendicular to the y-axis and passing through the point P (−4, 5) Proof Since the line is perpendicular to the y-axis, it is a vertical line with an undefined slope Its equation is of the form x = k, where k is a constant Since the line passes through P (−4, 5), we have x = −4 quod erat dem■ Question Determine the equation of the line perpendicular to the line y = 32 x + and passing through the point P (1, −4) Proof The given line has a slope of 32 The slope of a line perpendicular to it is the negative reciprocal, which is − 23 Using the point-slope form, the equation of the perpendicular line passing through P (1, −4) is y − (−4) = − 23 (x − 1) Simplifying, we get y = − 23 x − 10 quod erat dem■ Question Find the coordinates of the intersection point of the lines y = 2x + and y = −3x + Proof To find the intersection point, we need to solve the system of equations formed by the two lines Equating the expressions for y, we have 2x+1 = −3x+4 Solving this equation, we get x = 35 Substituting ( ) this value back into either equation, we find y = 35 + = 13 Therefore, the intersection point is ( 13 , 5 ) quod erat dem■ Question Determine the equation of the circle with center C(2, −1) and radius r = Proof The equation of a circle with center (h, k) and radius r is given by (x − h)2 + (y − k)2 = r2 Substituting the given values, we have (x − 2)2 + (y − (−1))2 = 42 Simplifying, we get (x − 2)2 + (y + 1)2 = 16 quod erat dem■ Question 10 Find the distance between the point P (3, 2) and the line 2x − 3y + = Proof The distance between a point (x0 , y0 ) and a line Ax + By + C = is given by the formula d = |Ax√0 +By0 +C| √ √2 Plugging in the values, we have d = |2(3)−3(2)+5| = |6−6+5| = √513 quod 4+9 A2 +B 2 +(−3) erat dem■ Question 11 Determine the equation of the parabola with vertex V (2, −3) and focus F (2, −5) Proof The standard equation of a parabola with vertex (h, k) and focus (h, k − 4a ) is given by (x − h)2 = 4a(y − k) In this case, the vertex is V (2, −3) and the focus is F (2, −5) The value of h remains the same for both the vertex and focus, so the equation becomes (x − 2)2 = 4a(y + 3) To find the value of a, we can use the distance formula between the vertex and focus The distance 1 between the vertex and focus is equal to 4a , so we have 4a = | − − (−3)| Solving for a, we get a = 18 ( ) Substituting this value back into the equation, we have (x − 2)2 = 81 (y + 3) Simplifying further, we get (x − 2)2 = 21 (y + 3) quod erat dem■ Question 12 Find the equation of the ellipse with center C(1, −2), major axis of length 10, and minor axis of length Proof The standard equation of an ellipse with center (h, k), major axis length 2a, and minor axis length 2 2b is given by (x−h) + (y−k) = In this case, the center is C(1, −2), the major axis length is 10, and the a2 b2 minor axis length is The value of h and k are the same as the coordinates of the center, so the equation becomes (x−1) + a2 (y+2)2 = b2 To find the values of a and b, we can use the lengths of the major and minor axes We have 2a = 10, so a = 5, and 2b = 6, so b = (y+2)2 Substituting these values back into the equation, we have (x−1) + = quod 32 erat dem■ Question 13 Find the equation of the hyperbola with center C(0, 0), transverse axis of length 8, and conjugate axis of length Proof The standard equation of a hyperbola with center (h, k), transverse axis length 2a, and conjugate 2 axis length 2b is given by xa2 − yb2 = In this case, the center is C(0, 0), the transverse axis length is 8, and the conjugate axis length is 2 The value of h and k are the same as the coordinates of the center, so the equation becomes xa2 − yb2 = To find the values of a and b, we can use the lengths of the transverse and conjugate axes We have 2a = 8, so a = 4, and 2b = 6, so b = 2 Substituting these values back into the equation, we have x42 − y32 = quod erat dem■ Question 14 Find the equation of the circle with center C(−2, 3) and radius Proof The standard equation of a circle with center (h, k) and radius r is given by (x − h)2 + (y − k)2 = r2 In this case, the center is C(−2, 3) and the radius is Substituting these values into the equation, we have (x + 2)2 + (y − 3)2 = 52 , which simplifies to (x + 2)2 + (y − 3)2 = 25 quod erat dem■ Question 15 Find the equation of the line passing through the points P (2, 5) and Q(−3, −1) Proof The equation of a line passing through two points (x1 , y1 ) and (x2 , y2 ) can be found using the −y1 (x − x1 ) point-slope form: y − y1 = xy22 −x In this case, the points are P (2, 5) and Q(−3, −1) Substituting the values into the equation, we have y − = −1−5 (x − 2) −3−2 (x − 2), which becomes y − = 65 (2 − x) Simplifying further, we get y − = −6 −5 Expanding and rearranging, we have y − = 12 − 65 x, which simplifies to y = 65 x + 15 quod erat dem■ Question 16 Find the equation of the line perpendicular to the line 3x + 4y = and passing through the point P (2, −1) Proof The given line has the form Ax + By = C, where A = 3, B = 4, and C = To find the perpendicular line, we need to swap the coefficients A and B and change the sign of one of them So, the equation of the perpendicular line will be 4x − 3y = D To find the value of D, we can substitute the coordinates of point P (2, −1) into the equation We have 4(2) − 3(−1) = D, which simplifies to + = D Therefore, D = 11 The equation of the perpendicular line is 4x − 3y = 11 quod erat dem■ Question 17 Find the equation of the line parallel to the line 2x − 5y = and passing through the point P (−1, 4) Proof The given line has the form Ax + By = C, where A = 2, B = −5, and C = To find a parallel line, we need to keep the coefficients A and B unchanged So, the equation of the parallel line will also be 2x − 5y = D To find the value of D, we can substitute the coordinates of point P (−1, 4) into the equation We have 2(−1) − 5(4) = D, which simplifies to −2 − 20 = D Therefore, D = −22 The equation of the parallel line is 2x − 5y = −22 quod erat dem■ Question 18 Find the equation of the line passing through the points P (3, 2) and Q(−2, 6) Proof The equation of a line passing through two points (x1 , y1 ) and (x2 , y2 ) can be found using the −y1 point-slope form: y − y1 = xy22 −x (x − x1 ) In this case, the points are P (3, 2) and Q(−2, 6) Substituting the values into the equation, we have 6−2 y − = −2−3 (x − 3) Simplifying further, we get y − = −5 (x − 3), which becomes y − = − 45 (3 − x) , which simplifies to y = − 54 x + 22 Expanding and rearranging, we have y − = − 45 x + 12 quod 5 erat dem■ Question 19 Find the equation of the line perpendicular to the line y = 23 x + and passing through the point P (2, −1) Proof The given line is in slope-intercept form y = mx + b, where m = 32 is the slope The perpendicular line will have a slope that is the negative reciprocal of 32 So, the slope of the perpendicular line is − 23 Using the point-slope form, we have y − (−1) = − 23 (x − 2) Simplifying further, we get y + = − 23 x + 43 Rearranging the equation, we have y = − 23 x + 13 quod erat dem■ Question 20 Find the equation of the line parallel to the line y = 2x − and passing through the point P (4, 5) Proof The given line is in slope-intercept form y = mx + b, where m = is the slope To find a parallel line, we need to keep the slope unchanged So, the equation of the parallel line will also be y = 2x + c To find the value of c, we can substitute the coordinates of point P (4, 5) into the equation We have = 2(4) + c, which simplifies to = + c Therefore, c = −3 The equation of the parallel line is y = 2x − quod erat dem■ Question 21 Find the equation of the line passing through the points P (1, 2) and Q(3, 6) Proof The equation of a line passing through two points (x1 , y1 ) and (x2 , y2 ) can be found using the −y1 (x − x1 ) point-slope form: y − y1 = xy22 −x In this case, the points are P (1, 2) and Q(3, 6) Substituting the values into the equation, we have y − = 6−2 (x − 1) 3−1 Simplifying further, we get y − = 42 (x − 1), which becomes y − = 2(x − 1) Expanding and rearranging, we have y − = 2x − 2, which simplifies to y = 2x quod erat dem■ Question 22 Find the equation of the line perpendicular to the line 4x − 5y = and passing through the point P (2, −1) Proof The given line has the form Ax + By = C, where A = 4, B = −5, and C = To find a perpendicular line, we need to swap the coefficients A and B and change the sign of one of them So, the equation of the perpendicular line will be −5x − 4y = D To find the value of D, we can substitute the coordinates of point P (2, −1) into the equation We have −5(2) − 4(−1) = D, which simplifies to −10 + = D Therefore, D = −6 The equation of the perpendicular line is −5x − 4y = −6 quod erat dem■ Question 23 Find the equation of the line parallel to the line 3x + 2y = and passing through the point P (−1, 4) Proof The given line has the form Ax + By = C, where A = 3, B = 2, and C = To find a parallel line, we need to keep the coefficients A and B unchanged So, the equation of the parallel line will also be 3x + 2y = D To find the value of D, we can substitute the coordinates of point P (−1, 4) into the equation We have 3(−1) + 2(4) = D, which simplifies to −3 + = D Therefore, D = The equation of the parallel line is 3x + 2y = quod erat dem■ Question 24 Find the equation of the line passing through the points P (−2, 3) and Q(4, −1) Proof The equation of a line passing through two points (x1 , y1 ) and (x2 , y2 ) can be found using the −y1 point-slope form: y − y1 = xy22 −x (x − x1 ) In this case, the points are P (−2, 3) and Q(4, −1) Substituting the values into the equation, we have −1−3 y − = 4−(−2) (x − (−2)) Simplifying further, we get y − = −4 (x + 2), which becomes y − = − 23 (x + 2) Expanding and rearranging, we have y − = − 23 x − 43 , which simplifies to y = − 32 x + 53 quod erat dem■ Mathematics