1. Grant, Liam, and Kody presents….

2. Slope Form of a Line – Equation of a Circle – Midpoint Formula – Distance Formula - d= Point slope form of a line- Standard form of a line – How to find slope- // lines have same slope Perpendicular lines- Vectors- (change in x, change in y)

3.  Find the distance between the points (-2,3) and (3,-2) and the slope.  Use the distance formula.  2 2 Use the slope formula.

4.  Given with this equation:  The center of the circle is (-7, 8)  The radius is 6/5. ◦ You can find 4 easy points by going each direction that lie on a solid x or y axis. ◦ Two find two more, we know that by looking at the graph, -6 on the x-axis goes through the circle twice. So we can plug in the -6 in our equation and solve for y. The second time, we can use -8 with the same concept. 2 2 2 2

5.  R(4,3); S(-3,6); T(2,1) Directions: Find the lengths of all three sides of triangle RST. Use the converse of the Pythagorean theorem to show that triangle RST is a right triangle. Show two of the sides have perpendicular slopes by using the products of the slopes.  Use the distance formula to find the length of each side, then use the converse of the Pythagorean Theorem. x*x + y*y = z*z 50 + 8 = 58  Find the product of the slope of ST and RT.  Since the product = -1 the lines are perpendicular

6.  Find k when vectors (8,k) and (9,6) are perpendicular.  Because the first coordinate is on the x axis and the second on the y axis, we can use the perpendicular slope equation. *These vectors are being acted as slopes.

7.  (7,2) + 3(-1,0) These are vectors. Plot them and solve.  The second term, you have solve it with multiplication, so it is (7,2)+(-3,0)  First go over 7, and then up 2.  Second go over -3 and do not go up or down.

8.  (1,1) and (4,7) Directions: find the equation of a line through these points.  Find the slope of the two points.  Plug the slope into the formula for a line using slope intercept form. y=2x+b  Plug in one of the points for the x and y variables. (4, 7)  Solve for b.  Now that you know the amount for b we can complete the equation.

9.  With point (7,-3) find the equation of a vertical line through this point. X can only be 7.  With point (7, -3) find an equation of a horizontal line through this point. Y can only be -3

10.  Find the equation of the line through the given point (5,7) parallel to the line. First, Put it into y-int. form to get slope.  =  Take out the y- intercept and plug in the (5,7) with the same slope (3) because they are //. Then solve for b.  To find the equation for a perpendicular line, plug in the x and the y and the reciprocal of the slope and solve for b.

11.  Given Q= (-1,-2) T= (7,4) S= (10,1) R= (6,-2) prove that QRST is a trapezoid.  Goal: ◦ First, prove the slope of QT is equal to the slope of RS.  Use the slope formula to find the slope of line QT.  Use it another time to the find the slope of line RS.  These slopes are equal, so the lines are parallel, which trapezoids have to have (bases parallel) ◦ Second, prove that the slope of TS does not equal QR.  Use the slope formula to find the slope of line QR and TS  These slopes are not equal, so the lines are not parallel, which trapezoids have to have.

12.  Given: ORQP is a parallelogram. R=(b, c) Q=(?, ?) P=(a, 0) O=(0, 0). M is the midpoint of RQ and N is the midpoint of OP. ◦ Q’s x-coordinate is ___ because you add the a, and drop a perp. down from R. The point where it meets line OP is b. Because it is a parallelogram, you can add triangle ORN onto the other side. The x- coordinate of Q is a+b. ◦ Q’s y-coordinate is c because the opposite lines are parallel.  Goal: Prove ONQM is a parallelogram. ◦ First, find what M is, by using the midpoint formula. M= (b+ ½ a, c) ◦ Then we find out what N is by doing the same exact thing. N= (½ a, 0) ◦ Prove OM(left)= NQ(right) by using the distance formula for each line. ◦ 2 2 2 2 ◦ After that, prove lines OM(top) and NQ(bottom) parallel by using the slope formula.