Math Distance Midpoint Notes Solutions

University:
Cornell University
Course:
MATH 1101 | Calculus Preparation
Academic year:

2020
Views:

103

Pages:

4
Author:

CometConductor

Distance, Midpoint, Lines (y = mx + b) Example 1: Finding the distance between two points Graph the points (-1, 4) and (3, -6) Draw a line segment connecting the two points Create a right triangle, using the segment you just drew as the hypotenuse Find the lengths of the sides of the right triangle Use the Pythagorean theorem to find the hypotenuse Distance Formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2) Example 2: Using the distance formula, find the following: a. Distance between (-1, 4) and (3, -6) d = sqrt((3 - (-1))^2 + (-6 - 4)^2) = sqrt(16 + 100) = sqrt(116) = 2sqrt(29) b. Distance between (c, d) and (-c, -d) d = sqrt((-c - c)^2 + (-d - d)^2) = sqrt(4c^2 + 4d^2) = 2sqrt(c^2 + d^2) Midpoint Formula: ((x1 + x2)/2, (y1 + y2)/2) Example 3: Find the midpoint of the line segment with the given endpoints a. (-1, 4) and (3, -6) ((-1 + 3)/2, (4 + (-6))/2) -> (1, -1) b. (2x, 3x) and (-5x, -4x) ((2x + (-5x))/2, (3x + (-4x))/2) -> (-3x/2, -x/2) Slope-Intercept Form of a Line y = mx + b m = Slope (0, b) = y-intercept Need to get "y" alone! y-intercept: When x = 0, y is the y-intercept. Example 4: Find the slope and y-intercept of the given line a. y = (1/2)x - 4 m = 1/2 b = -4 (0, -4) b. 3y = -6x + 3 y = -2x + 1 m = -2 b = 3 (0, 3) c. 4x - 3y = 7 -3y = -4x + 7 y = (4/3)x - 7/3 m = 4/3 b = -7/3 (0, -7/3) Example 5: Graph the equation of the line. If the line is not in slope-intercept form, put it in that form first, then use the y-intercept and slope to graph the line. a. 1 y = (7/3)x - 1 m = 7/3 (0, -1) b. (1/2)y = -x + 3 y = -2x + 6 m = -2 (0, 6) c. 4x - 3y = 12 -3y = -4x + 12 y = (4/3)x - 4 m = 4/3 (0, -4) Distance, Midpoint, Lines (y = mx + b) For all distance and midpoint problems, please write your answer on the answer line provided. For all equations of lines, please draw your lines on the graphs provided. Find the distance between the points whose coordinates are given. (6, 4) and (-8, 11) √((-8-6)²+(11-4)²) = √(196+49) = √245 = 7√5 (-5, 8) and (-10, 14) √((-10+5)²+(14-8)²) = √(25+36) = √61 (a, b) and (-a, -b) √((-a-a)²+(-b-b)²) = √(4a²+4b²) = 2√(a²+b²) (x, 4x) and (-2x, 3x) Given x > 0 √((-2x-x)²+(3x-4x)²) = √(9x²+x²) = √10x² = x√10 Find the midpoint of the line segment with the given endpoints. (1, -1) and (5, 5) ((1+5)/2, (-1+5)/2) = (3, 2) (6, -3) and (6, 11) ((6+6)/2, (-3+11)/2) = (6, 4) (1.75, 2.25) and (-3.5, 5.57) ((1.75-3.5)/2, (2.25+5.57)/2) = (-0.875, 3.91) (x, 4x) and (-2x, 3x) ((x-2x)/2, (4x+3x)/2) = (-x/2, 7x/2)