Problems All Star Team : Marcus Dennis, Wyatt Dillon, Will Clansky

Find the distance between these two points and the slope: (-2,3) and (3,-2) Distance between points (-2, 3) and (3,-2): Use Distance formula: Slope of the line that goes through these two points : Use formula to find the slope of a line: 39)

40) Graph this circle, indicating the center, radius, and 6 points on the circle: 4 of the 6 points: -7, , , , 8 6 points on the circle: Center of Circle: (-7,8) Radius: Other two points: Two more Points:

Find the lengths of all three sides of RST. Use the converse of the Pythagorean Theorem to show that RST is a right triangle. Show two of the sides have perpendicular slopes by using the products of the slopes. 41) Converse of Pythagorean Theorem: If A +B =C then the triangle is a right triangle

41) (CONT) Two lines are perpendicular when the product of their slopes is equal to -1 SLOPE:

The two vectors are perpendicular, find the value of K: (8,k) and (9,6) 42) Two lines are perpendicular when the product of their slopes is equal to -1

43). A) Plot each B) Solve (7,2) + (-3,0)= (4,2) Vectors can be added by the rule: (a,b) + (c,d) = (a+c,b+d)

44). Find the equation of a line through these points SLOPE =7-1 / 6-3= 6/3 = 2 Y-1= 2(x-1) A line that passes through point (x,y) and has slope m = y-y1= m(x-x1) ( 1,1) and (4,7)

45)a) Find the equation of a vertical line through this point b) Find the equation of a horizontal line through this point : A) x=7 B) y+3= 0(x-7) y=-3 (7,-3) Y= -3 X=7 Slope of a vertical line is not defined The slope of a horizontal line is 0

46).A) Find the equation of the line through the given point parallel to the line. B) Find the equation of the line through the given point parallel to the line. -2y= -6x + 5 Y= 3x- 5/2 a)y-7= 3(x-5) b)Y-7= -1/3(x-5) Point slope form= y- y1= m(x-x1) Y=mx+b

a. Show that QRST is a trapezoid using the numeric coordinates given -Trapezoid – two parallel sides -Slope is rise over run -Congruent slopes means parallel lines -Slope QT = 6/8 = ¾ -Slope RS = ¾ ¾ = ¾ T (7,4) S (10,1) R (6,-2)Q (-1,-2) b. Find the equation of the altitude to line QT through R 47. -Altitude – A perpendicular line from a point through the given line -Q = (6,-2) -Perpendicular lines have slopes that when multiplied equal -1 -Slope of line QT is 3/4, so altitude R must have slope -4/3 -With the slope and one set of coordinates, we can find the equation: (y – (-2) = -4/3(x – 6) Or (y + 2) = -4/3(x – 6)

48. Given the parallelogram shown. Let M Be the midpoint of line RQ and N be the Midpoint of line OP. Use coordinate Geometry to prove that ONQM is a Parallelogram. Q (a+b, c) Because opposite sides of a parallelogram Are congruent, the distance of RQ is the Same distance of OP, which is (distance a). Therefore, the distance of Q on the x-axis Is the same as (distance a) + (distance b) or (a+b). Parallelogram – opposite sides congruent Slopes are congruent Midpoint line – ([x1 + x2/2], [y1 + y2]/2) don’t need to worry about, though Because we have proved that RQ and OP are congruent segments, they will also Form congruent midpoints, which in turn make two congruent segments, meaning segment ON = NP = RM = MQ The slope of OM is b/(R+ 1/2distance a) The slope of NQ is also b/(1/2distance a) Because ON = MQ and OM and NQ have congruent slopes, ONQM is a parallelogram N M Distance b Distance a

ax + by = c Distance formula for lines on a grid Equation to find a midpoint of a line on a grid Y intercept of a line on a graph Equation for a circle on a graph with one coodinate Slope formula Point slope formula with one coodinate Standard formula of a line on a graph