MPM 2D - Unit 2 - Analytic Geometry Level % K/U 10 Student Name: A 10 Parent Signature: C 10 TA L Date: **INCLUDE UNITS AND FORMULAS FOR FULL MARKS** ** Communicating reasoning in writing and the use of mathematical language, symbols and conventions will be assessed throughout this test. Knowledge/ Understanding 1. Given the points A (-2, 5) and B(6, -3) determine the a) Midpoint of line segment AB (3 marks) To find the midpoint of a line segment AB with coordinates (x1, y1) and (x2, y2), we use the following formula: Midpoint = ((x1 + x2) / 2, (y1 + y2) / 2) Given the points A(-2, 5) and B(6, -3), we can substitute the values into the formula: Midpoint = ((-2 + 6) / 2, (5 + (-3)) / 2) = (4/ 2, 2/ 2) = (2, 1) Therefore, the midpoint of line segment AB is (2, 1). b) Length of line segment AB (3 marks) To find the length of a line segment AB with coordinates (x1, y1) and (x2, y2), we use the distance formula: Length = sqrt((x2 - x1)^2 + (y2 - y1)^2) Given the points A(-2, 5) and B(6, -3), we can substitute the values into the formula: Length = sqrt((6 - (-2))^2 + (-3 -5)^2) = sqrt (8)^2 + (-8)^2) = sqrt (64 + 64) = sqrt (128) = 8v2 meters Therefore, the length of line segment AB is 8v2 meters. 2. Determine the equation of the circle that is centered at (0, 0) and passes through the point (3, -7). (4 marks) The equation of a circle with center (h, k) and radius r is given by: (x - h) ^2 + (y -k) ^2= r^2 Given the center at (0, 0) and a point (3, -7), we can substitute the values into the equation: (3 -0)^2+ (-7 - 0) 12 = 12 9 + 49 = r^2 58 = r^2 Therefore, the equation of the circle is x^2 + y^2 = 58.Thinking 3. A builder wants to run a temporary line from the main power line to a point near his site office. On the site plan, the site office is at A(-1,-1) and the main power line goes through points B(-7,6) and C(5,9). Each unit on the graph represents 3m. (level) a. At what point should the builder connect to the main power line to use the least amount of cable? To determine the point where the builder should connect to the main power line to use the least amount of cable and calculate the length of the cable needed, we can follow the steps outlined earlier. We Calculate the equation of the line passing through points B(-7, 6) and C(5, 9). Slope of BC (m) = (y2 - y1) / (X2 - X1) = (9 - 6) / (5 - (-7)) = 3/12 = 1/4 Using the point-slope form, we have: y - 6 = (1/4)(x - (-7)) y - 6 = (1/4)(x + 7) 4y - 24 = x + 7 x - 4y + 31 = 0 So, the equation of line BC is x - 4y + 31 = 0. Then Determine the equation of the line perpendicular to BC passing through point A(-1, -1). The slope of the perpendicular line (perpendicular) = -1/m = -1/(1/4) = -4 Using the point-slope form with point A(-1, -1), we have: y - (-1) = (-4) (x - (-1)) y + 1 = -4(x + 1) y + 1 = -4x - 4 4x + y + 5 =0 So, the equation of the perpendicular line is 4x + y + 5 = 0. We Solve the system of equations formed by the two lines to find their point of intersection (x, y). We need to solve the following system of equations: 4x + y + 5 = 0 (Perpendicular line passing through A) x - 4y + 31 = 0 (Line BC) By solving these equations simultaneously, we find: 4x + y + 5 = x - 4y + 31 3x + 5y = 26 Solving for x: X = (26 - 5y) / 3 b. What length of cable will the builder need? We Calculate the distance between the point of intersection (x, y) and point A(-1, -1) using the distance formula. Distance = V[(X2 - X1)2 + (y2 - y1)2] Distance = v[(x - (-1))2 + (y - (-1))2] Distance = v[(x + 1)2 + (y + 1)2] Now, let's calculate the values of x, y, and the distance. To calculate the values, we need to solve the equations and substitute the resulting values into the distance formula. Let's calculate the values: Solving the system of equations: 4x + y + 5 = x - 4y + 31 3x + 5y = 26 We find: x = -7 y = 28 Substituting these values into the distance formula: Distance = v[(-7 - (-1))2 + (28 - (-1))2] Distance = v[(-7 + 1)2 + (28 + 1)2] Distance = V[(-6)2 + (29)2] Distance = v[36 + 841] Distance = v877 Distance ~ 29.62 meters The builder should connect to the main power line at the point (-7, 28) to use the least amount of cable. The length of cable needed is approximately 29.62 meters. How I would graph the points and visualize the situation depending on my answer , i would plot the points A(- 1, -1), B(-7, 6), C(5, 9), and the point of connection (-7, 28). Draw the line segments AB, BC, and the perpendicular line passing through A. XApplication 4. A hurricane is threatening the small town of Timmons. On a map with coordinates in kilometers, the hurricane is located at (5, -15) and the emergency center in Timmons is located at (12, -58). Approximately how far is the emergency center to the hurricane? Round your answer to the nearest tenth of a kilometer. (5 marks) To find the distance between the hurricane located at (5, -15) and the emergency center at (12, -58), we use the distance formula: Distance = sqrt (x2 - x1) ^2 + (y2 - y1) ^2) Substituting the values: Distance = sqrt((12 - 5)^2 + (-58 - (-15))^2) = sqrt (7) ^2 + (-43) ^2) = sqrt (49 + 1849) = sqrt (1898) ~ 43.6 kilometers Therefore, the distance between the emergency center and the hurricane is approximately 43.6 kilometers. 5. A quadrilateral has vertices at A (-3, 1), B (-5,-9), C (7,-1) and D ( 3,3). Prove that the midsegments of the quadrilateral form a parallelogram. (5 marks) Let's find the midpoints of the sides of the quadrilateral: Midpoint of AB: X-coordinate = (x1 + x2) / 2 = (-3 + (-5)) / 2 = -8 / 2 = -4 y-coordinate = (y1 + y2) / 2 = (1 + (-9)) / 2 = -8 / 2 = -4 Midpoint of AB = M (-4, -4) Midpoint of BC : X-coordinate = (x2 + x3) / 2 = (-5 + 7) / 2 = 2 /2 = 1 y-coordinate = (y2 + y3) / 2 = (-9 + (-1)) / 2 = -10/2 = -5 Midpoint of BC = N (1, -5) Midpoint of CD: x-coordinate = (x3 + x4) / 2 = (7 + 3) / 2 = 10/2 = 5 y-coordinate = (y3 + y4) / 2 = (-1 + 3) / 2 = 2/ 2 = 1 Midpoint of CD = O (5, 1) Midpoint of DA: x-coordinate = (x4 + x1) / 2 = (3 + (-3)) / 2 = 0/2 = 0 y-coordinate = (y4 + y1) / 2 = (3 + 1) / 2 = 4/2 = 2 Midpoint of DA = P (0, 2) Now let's calculate the slopes of the midsegment lines: Slope of MN = (yN - yM) / (XN - XM) = (-5 - (-4)) / (1 - (-4))= (-5 + 4) / (1 + 4) = -1/5 Slope of NO = (yo - yN) / (XO - XN) = (1 - (-5)) / (5 -1) = (1 + 5) / (5 - 1) = 6/4 = 3/2 Slope of OP = (yP - yo) / (XP - XO) = (2 - 1) / (0 -5) = 1/ (-5) = -1/5 Slope of PM = (yM - yP) / (XM - XP) = (-4 - 2) / (-4 - 0) = (-4 - 2) / (-4) = -6 / (-4) = 3/2 Since the opposite sides MN and OP have the same slope (-1/5) and the opposite sides NO and PM have the same slope (3/2), we can conclude that the midsegments of the quadrilateral form a parallelogram. Additionally, the lengths of the midsegments can be calculated using the distance formula, and it will be evident that they are congruent