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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