{ Geometry Semester 1 Review. There are two ways to do graph an equation that is listed in STANDARD...

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{ Geometry Semester 1 Review

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 There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept form. In this case it would be: y = -2/3 x Graph 2x + 3y = 6 2. Most efficient way to graph would be to find the x and y intercepts by plugging in 0 for the x and solve for y, next repeat but plug in 0 for y and solve for x. In this case 2(0)+3y=6, y=2 and 2x+3(0)=6, x = 3. Now just graph those points and connect the dots.

Transcript of { Geometry Semester 1 Review. There are two ways to do graph an equation that is listed in STANDARD...

Page 1: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

{Geometry

Semester 1 Review

Page 2: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.
Page 3: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

There are two ways to do graph an equation that is listed in STANDARD FORM.

1. Rearrange the equation into y-intercept form. In this case it would be: y = -2/3 x + 2

2. Graph 2x + 3y = 6

2. Most efficient way to graph would be to find the x and y intercepts by plugging in 0 for the x and solve for y, next repeat but plug in 0 for y and solve for x. In this case 2(0)+3y=6, y=2 and 2x+3(0)=6, x = 3. Now just graph those points and connect the dots.

Page 4: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

Step one is to find the slope of our given equation.

3. Write an equation that is perpendicular to 4x + 8y = -36 and passes through the point (0, 7)

4x + 8y = -36-4x -4x 8y = -4x – 36 8 8 8 y = -1/2 x – 4.5

Step two is to find the slope of the new equation. Remember that parallel lines have the same slope while perpendicular lines slopes are opposite reciprocals. Therefore, the slope of our new equation is +2

Step three is to find the y-intercept. In this case it is easy since (0,7) was given to us and the y-intercept is the y value when x = 0.

ANSWER:y = 2x + 7

Page 5: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

4. What is the value of x and y?y = 52° (C-18) states that if a triangle is isosceles, then the base angles are congruent.

Finding x:We know that the interior angles in a triangle are 360°, therefore……

180 – 52 – 52 = 76

14x + 6 = 76 -6 -6 14x = 7014 14

x = 5

Page 6: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

5. The coordinates of three vertices of a quadrilateral are shown below. What would the ordered pair for the fourth vertex be to make this quadrilateral a parallelogram?

(4,9)

(1, 5)

(5, 5)

(x, y)

Step 1: since a parallelogram has to have two sets of parallel lines we need to know the slope of one line.

9 – 5 = 44 – 1 = 3

Slope = 4/3Step 2: Plug our unknown (x,y) into the slope equation and solve for them.

5 – y = 45 – x = 3

y = 1 and x = 2ANSWER: (2, 1)

Page 7: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

6. Line A is parallel to line B. What are the values of x and y?

(8x +14)°

(12x – 42)°

(y – 3)°

SOLUTION: Step 1: Solve for x.(12x – 42) and (8x + 14) are AIA and therefore congruent.

12x – 42 = 8x + 14 + 42 +42 12x = 8x + 56 -8x -8x 4x = 56 4 4 x = 14

STEP 2:(8x + 14) and (y – 3) are vertical angles which means they are congruent. Since we know x = 14 ---- 8(14) + 14 = 126

y – 3 = 126 + 3 +3 y = 129

Page 8: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

7. What geometric relationship does the model demonstrate.

ASA

Page 9: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

8. What is the equation of the line graphed below?

y = 2x - 5

Page 10: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

Remember the midpoint is an average of the x values and an average of the y values since it is directly in the middle of the two endpoints.

X coordinate: y coordinate:-2 + x = 3 7 + y = 8 2 2Multiply both sides by 2

-2 + x = 6 7 + y = 16+2 +2 -7 -7 x = 8 y = 9

9. What are the coordinates of point B if (3, 8) is the midpoint between points A (-2, 7)?

The coordinates of B are (8, 9)

Page 11: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

10. In the figure, ray AE is parallel to ray DF, and segment BD intersects AE at C. What is the measure of angle BAC?

B

A

D

C

Since AE and DF are parallel angle BCE corresponds to 45°

E

F45°

25°

45°135°

• Angle BCE (45°) is a linear pair with Angle BCA therefore BCA = 180 – 45 = 135°

We now know two out of the three angles in the triangle BAC. <BAC = 180 – 135 – 25

<BAC = 20°

Page 12: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

11. Which triangle congruence conjecture (theorem) can be use in the figure below?

ASA

Page 13: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

12. a. List all congruent angles in the parallelogram below. b. List all supplementary angles in the parallelogram. <A ≅ <C and <B

≅ <D<A + <D = 180°<A + <B = 180°<C + <D = 180°<C + <B = 180°

Page 14: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

x -5 -3 -1 1 3 5 7y 2 3 4 5 6 7 8

13. What is the equation of the line that fits the data below?

STEP 1:Find the slope by rise/run or change in y/change in xSlope = ½

STEP 2:Find the y-intercept (y value when x = 0). Midpoint between 4 and 5 or choose a point from the table and use point – slope form to solve.

y – 5 = ½(x – 1) then simplify

y = ½ x + 4 ½

Page 15: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

14. Which quadrilaterals have the property, diagonals are congruent?

Isosceles Trapezoid

Rectangle

Square

Page 16: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

Remember; midpoints are just the average of the x values and the average of the y values.

15. What are the coordinates of the midpoint of the segment connecting the points (-3, 8) and (11, -4)?

x – value: y – value:

-3 + 11 = 8 + -4 = 2 2

8 = 4 = 2 2

4 2

(4, 2) = midpoint

Page 17: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

16. An octagon has angle measures 100°, 85°, 130°, 155°. The remaining four angles are congruent. What are the measures of the remaining angles?STEP 1:Find the total sum of the interior angles using 180(n-2).180(8 – 2) = 180(6) = 1080

STEP 2:Subtract the known angle measures from 1080.1080 – (100 + 85 + 130 + 155) = 1080 – 470 = 610

STEP 3:Divide 610 by 4 since all four remaining angles are congruent.610 ÷ 4 = 152.5°

Page 18: { Geometry Semester 1 Review.  There are two ways to do graph an equation that is listed in STANDARD FORM.  1. Rearrange the equation into y-intercept.

17. Write an equation that represents a line that is parallel to the line 4x – 2y = 16 and passes through the point (5, 9)

STEP 1:Rearrange current equation from STANDARD form to Y-INTERCEPT form. 4x – 2y = 16-4x -4x -2y = -4x + 16 -2 -2 -2

y = 2x - 8

Parallel lines have the same slope; therefore, our new equation will have the slope of 2.

STEP 2:Use POINT – SLOPE FORM y –y1 = m(x – x1) m = 2, x1 = 5, y1 = 9

y – 9 = 2(x – 5)y – 9 = 2x – 10

+9 +9

y = 2x -1