Task 2- College Geometry

A. Use only the analytic methods included in the “Analytic Geometry Graph and Permitted Analytic Methods” attachment to do the following: 1. Compute the coordinates for point M , showing and explaining all work. To compute the coordinates for point M, you must use the midpoint formula to find the midpoint of KN . Point K is (1,1) and point N (2,1). To find the x-coordinate of point M, our midpoint, we must add both x-coordinates (1 for point K and 2 for point N) and then divide by 2. 1 + 2 = 3 and 3 ÷ 2 = 1.5 Thus, the x-coordinate of point M, our midpoint, will be 1.5. To find the y-coordinate of point M, our midpoint, we must add both y-coordinates (1 for point K and 1 for point N) and then divide by 2. 1 + 1 = 2 and 2 ÷ 2 = 1 Thus, the y- coordinate of point M, our midpoint, will be 1. By substituting these calculations into our formula, you can determine that the midpoint of line KN, which is point M, will be (1.5, 1). 2. Demonstrate that ∆ SKP is an isosceles right triangle by doing the following: a. Demonstrate that the triangle in part A2 is an isosceles triangle, showing and explaining all work. For the triangle SKP to be an isosceles right triangle, it must have two equal sides. We can determine if this is true by calculating the distance of each line/side of the triangle using the distance formula: . To calculate line SK, you must use point S (1,0) and point K (1,1). Substitute the coordinates of points s and K into the formula and further simplify to calculate the distance: √ ( 1 − 1 ) 2 +( 1 − 0 ) 2 Substitute x 1 = 1 , y 1 = 0 , x 2 = 1 , y 2 = 1 √ ( 0 ) 2 +( 1 ) 2 Calculate the parentheses next by subtracting the x and y coordinates. √ 0 + 1 Square the differences. √ 1 Add the squared values. 1 Take the square root of the sum. The distance of line SK is 1. To calculate line KP, you must use point K (1,1) and point P (1.5,0.5). Substitute the coordinates of points s and K into the formula and further simplify to calculate the distance: √ ( 1.5 − 1 ) 2 +( 0.5 − 1 ) 2 Substitute x 1 = 1 , y 1 = 1 , x 2 = 1.5 , y 2 = 0.5 √ ( 0.5 ) 2 +(− 0.5 ) 2 Calculate the parentheses next by subtracting the x and y coordinates. √ 0.25 + 0.25 Square the differences. √ 0.5 Add the squared values. ≈ 0.71 Take the square root of the sum. The distance of line KP is √ 0.5 or approximately 0.71.

To calculate line PS, you must use point P (1.5,0.5) and point S (1,0). Substitute the coordinates of points s and K into the formula and further simplify to calculate the distance: √ ( 1 − 1.5 ) 2 +( 0 − 0.5 ) 2 Substitute x 1 = 1.5 , y 1 = 0.5 , x 2 = 1 , y 2 = 0 √ (− 0.5 ) 2 +(− 0.5 ) 2 Calculate the parentheses next by subtracting the x and y coordinates. √ 0.25 + 0.25 Square the differences. √ 0.5 Add the squared values. ≈ 0.71 Take the square root of the sum. The distance of line PS is √ 0.5 or approximately 0.71. Since both Lines KP and PS are equal, triangle SKP is indeed an isosceles triangle. b. Demonstrate that the triangle in part A2 is a right triangle, showing and explaining all work. Using the Pythagorean Theorem, we know that if the square of the length of the longest side is equal to the sum of the squares of the other two sides, then the triangle is a right triangle. a 2 + b 2 = c 2 √ 0.5 2 + √ 0.5 2 = 1 2 Using the Pythagorean Theorem, substitute the side lengths from part A2. 0.5 + 0.5 = 1 Because our distances were √ 0.5 , when we square a square root, they cancel one another. 1 = 1 Add together the remaining 0.5 + 0.5. Its sum of 1 equals our longest side. This concludes that the triangle SKP is a right triangle. 3. Demonstrate that ☐ MNQP is a square by doing the following: a. Demonstrate that the quadrilateral in part A3 is a rhombus, showing and explaining all work. For ☐ MNQP to be a rhombus, each of its sides must be equal. We can demonstrate that this is true with ☐ MNQP by calculating the distance of each of its sides using the distance formula: To calculate line MN, you must use point M (1.5, 1) and point N (2, 1). Substitute the coordinates of points M and N into the formula and further simplify to calculate the distance: √ ( 2 − 1.5 ) 2 +( 1 − 1 ) 2 Substitute x 1 = 1.5 , y 1 = 1 , x 2 = 2 , y 2 = 1 √ ( 0.5 ) 2 +( 0 ) 2 Calculate the parentheses next by subtracting the x and y coordinates. √ 0.25 + 0 Square the differences. √ 0.25 Add the squared values. 0.5 Take the square root of the sum. The distance of line MN is 0.5.

( 1 − 1 ) ( 2 − 1.5 ) Substitute x 1 = 1.5 , y 1 = 1 , x 2 = 2 , y 2 = 1 ( 0 ) ( 0.5 ) Calculate the parentheses next by subtracting the x and y coordinates. 0 The slope of MN is 0. Making it a horizontal line. Slope of NQ: point N (2, 1) and point Q (2, 0.5). ( 0.5 − 1 ) ( 2 − 2 ) Substitute x 1 = 2 , y 1 = 1 , x 2 = 2 , y 2 = 0.5 (− 0.5 ) ( 0 ) Calculate the parentheses next by subtracting the x and y coordinates. undefined The slope of NQ is undefined. Making it a vertical line. Slope of QP: point Q (2, 0.5) and point P (1.5, 0.5). ( 0.5 − 0.5 ) ( 1.5 − 2 ) Substitute x 1 = 2 , y 1 = 0.5 , x 2 = 1.5 , y 2 = 0.5 ( 0 ) (− 0.5 ) Calculate the parentheses next by subtracting the x and y coordinates. 0 The slope of QP is 0. Making it a horizontal line. Slope of MP: point M (1.5, 1) and point P (1.5, 0.5). ( 0.5 − 1 ) ( 1.5 − 1.5 ) Substitute x 1 = 1.5 , y 1 = 1 , x 2 = 1.5 , y 2 = 0.5 (− 0.5 ) ( 0 ) Calculate the parentheses next by subtracting the x and y coordinates. undefined The slope of MP is undefined. Making it a vertical line. Since the slopes of MN and QP are both 0 making them horizontal lines and both NQ and MP are undefined making them vertical lines, this means that each pair of consecutive sides are perpendicular, proving that all angles are right angles. 4. Demonstrate that ∆ RKS is similar to ∆ SKP from part A2 by demonstrating that corresponding sides all have the same scale factor (i.e. are proportional), showing and explaining all work.

To find whether ∆ RKS and ∆ SKP are proportional, we must first calculate the length of each side using the distance formula: Side lengths of ∆ RKS: Line RK √ ( 1 − 0 ) 2 +( 1 − 0 ) 2 Substitute x 1 = 0 , y 1 = 0 , x 2 = 1 , y 2 = 1 √ ( 1 ) 2 +( 1 ) 2 Calculate the parentheses next by subtracting the x and y coordinates. √ 1 + 1 Square the differences. √ 2 Add the squared values. The distance of line RK is √ 2 . Line KS √ ( 1 − 1 ) 2 +( 0 − 1 ) 2 Substitute x 1 = 1 , y 1 = 1 , x 2 = 1 , y 2 = 0 √ ( 0 ) 2 +(− 1 ) 2 Calculate the parentheses next by subtracting the x and y coordinates. √ 0 + 1 Square the differences. √ 1 Add the squared values. The distance of line KS is √ 1 . Line SR √ ( 0 − 1 ) 2 +( 0 − 0 ) 2 Substitute x 1 = 1 , y 1 = 0 , x 2 = 0 , y 2 = 0 √ (− 1 ) 2 +( 0 ) 2 Calculate the parentheses next by subtracting the x and y coordinates. √ 1 + 0 Square the differences. √ 1 Add the squared values. The distance of line SR is √ 1 . Side lengths of ∆ SKP: Line SK √ ( 1 − 1 ) 2 +( 1 − 0 ) 2 Substitute x 1 = 1 , y 1 = 0 , x 2 = 1 , y 2 = 1 √ ( 0 ) 2 +( 1 ) 2 Calculate the parentheses next by subtracting the x and y coordinates. √ 0 + 1 Square the differences. √ 1 Add the squared values. The distance of line SK is √ 1 . Line KP √ ( 1.5 − 1 ) 2 +( 0.5 − 1 ) 2 Substitute x 1 = 1 , y 1 = 1 , x 2 = 1.5 , y 2 = 0.5 √ ( 0.5 ) 2 +(− 0.5 ) 2 Calculate the parentheses next by subtracting the x and y coordinates. √ 0.25 + 0.25 Square the differences. √ 0.5 Add the squared values. The distance of line KP is √ 0.5 .