Coordinate Geometry Unit Test Review

 Math 2: Algebra 2, Geometry and Statistics
Ms. Sheppard-Brick 617.596.4133
http://lps.lexingtonma.org/Page/2434
Coordinate Geometry Unit Test Review Name:
Date:
1. Given the points A(1, 3) and B(7, –5) a. Find the midpoint. b. Find the distance between them. c. Find the equation of the line that goes through both points. d. Find the equation of the line that is parallel to that line through C(3, 4). e. Find the coordinates of a fourth point that would form a trapezoid with A, B, and C. f. Find the equation of the line that is perpendicular to AB through E(3, –3). g. Find the distance of point E to the line AB. h. Find the area of triangle ABE. i. Find the coordinates of a fourth point that would form a parallelogram with A, B, and E. j. Find the equation of a circle that has A as its center and B is on the circle. You should be able to find more than 5 more points on this circle. k. Find the equations of the two horizontal and two vertical tangents to the circle. l. Find the equation of a tangent to the circle at point B. m. Find the set of points that are the same distance from A and B. n. Find the point(s) of intersection between the line through AB and 2x – y = 6. o. Find the point(s) of intersection between the line through AB and (x – 2)2 + (y + 1)2 = 9. 2. Given the coordinates Q(–3, –2), U(–6, 5), A(1, 8), D(4, 1) a. Graph the coordinates. What type of special quadrilateral does this appear to be? b. What are the properties of this type of quadrilateral? c. What, therefore, must you show is true? d. Prove what kind of special quadrilateral this is. 3. a. Put a trapezoid on a coordinate grid. Label the vertices T, R, A, P and assign variables to the coordinates of the vertices. (Note 1: you will be finding midpoints in part c, so there may be a way of assigning variables that makes them easier to work with. Note 2: Make sure the trapezoid is not isosceles.) b. Find the lengths of the bases (the parallel sides) then average them. c. Find the midpoints of the legs (the non-­‐parallel sides). Label them M and N. d. Find the length of MN. e. What do you notices about the length of MN and the average lengths of the bases? f. Find the midpoints of the diagonals. Do the diagonals of a general trapezoid bisect each other? g. What would you need to do to your variables to make TRAP an isosceles trapezoid? (An isosceles trapezoid is one where the non-­‐parallel sides have the same length.) h. Do the diagonals of an isosceles trapezoid bisect each other? Math 2: Algebra 2, Geometry and Statistics
Ms. Sheppard-Brick 617.596.4133
http://lps.lexingtonma.org/Page/2434
Name:
Date:
Answer Key: 4
#1+ 7 3 − 5 &
1. a. midpoint M = %
,
( = ( 4, −1) $ 2
2 '
b. d = (7 −1) 2 + (−5 − 3) 2 = 6 2 + (−8) 2 = 100 = 10 €
−5 − 3 −8
4
c. m =
=
= − so the equation is y − 3 = − 43 (x −1) 7 −1
6
3
€
or y + 5 = − 43 (x − 7) C
4
A
d. y − 4 = − 43 (x − 3) €2
€
€
e. any point on the line in d, 5
for example (6, 0) €
–2
3
f. y + 3 = 4 (x − 3) –4
g. intersection point of AB B
–6
and the perpendicular line €
is (4.6, –1.8) so the distance between point E and the line is d = (4.6 − 3) 2 + (−1.8 + 3) 2 = 1.6 2 +1.2 2 = 4 = 2 h. Area = 10 €
i. From E to B is down 2, right 4 so if you go down 2, right 4 from A, you get the point (5, 1). Alternatively, E is down 6, right 2 from A so if you go down 6, right 2 from B, you get (9, –11). Are there other points that work? x = -9
2
2
2
j. (x – 1) + (y – 3) = 10 , some other points are (1, 13), (1, –7), (11, 3), (–9, 3), (7, 11), (–5, –5), (–5, 11) k. Horizontal tangents: y = 13, y = –7 Vertical tangents: x = 11, x = –9 l. y + 5 = 43 (x − 7) – 10
A
2
5
10
M(4, -1)
–2
–4
B
–6
10
4
A
2
5
10
–2
–4
E
B
–6
14
y = 13
12
x = 11
10
8
6
4
A
2
–5
5
10
–2
€
–4
y = -7
–6
–8
y +5=.75(x-7)
B
Math 2: Algebra 2, Geometry and Statistics
Ms. Sheppard-Brick 617.596.4133
http://lps.lexingtonma.org/Page/2434
€
m. this locus is the perpendicular bisector of AB: y +1 = 43 (x − 4) n. The lines intersect at (3.1, 0.2) o. The circle intersects the line at (1.757, 1.990) and (4.803, –2.070) (use your calculator for this secant problem) 4
Name:
Date:
A
2
5
–2
M
–4
B
–6
4
A
2x-y=6
4
2
A
(1.757, 1.990)
2
(3.1, 0.2)
5
10
–2
5
10
–2
(4.803, -2.070)
–4
–4
B
–6
B
–6
2. a. The figure appears to be a square 8
A
6
b. All the sides are the same length, opposite sides have the U
same slope, the slopes are perpendicular at the vertices, 4
the diagonals are the same length. 2
c. If the sides are shown to be the same length and the slopes are all perpendicular at all the vertices, it is proved –5
to be a square. –2
Q
d. dQU = (−3 + 6) 2 + (−2 − 5) 2 = 32 + (−7) 2 = 58 dUA = (−6 −1) 2 + (5 − 8) 2 = (−7) 2 + (−3) 2 = 58 dAD = (1 − 4) 2 + (8 −1) 2 = (−3) 2 + 7 2 = 58 €
€
€
€
€
10
dDQ = (4 + 3) 2 + (1+ 2) 2 = 7 2 + 32 = 58 so all of the sides are the same length −2 − 5
7
5 − 8 −3 3
mQU =
= − mUA =
=
= −3 + 6
3
−6 −1 −7 7
8 −1
7
1+ 2 3
so the slopes are perpendicular at each mAD =
= − mDQ =
= 1− 4
3
4+3 7
vertex €
D
5
Math 2: Algebra 2, Geometry and Statistics
Ms. Sheppard-Brick 617.596.4133
http://lps.lexingtonma.org/Page/2434
3. a. You can choose to double the variables so the midpoints don’t have fractions. b. dTP = 2a, dRA = 2d – 2b, the average of them is d – b + a. c. midpoints M(b, c) and N(a+d, c) d. dMN = a + d – b e. the average and the length of MN are the same. f. The midpoints (d, c) and (a + b, c) are not the same point so the diagonals do not bisect each other. g. Isosceles trapezoids have non-­‐
parallel sides with opposite slopes so point A has the coordinates (2a – 2b, 2c) because point R is right 2b, up 2c from point T so point A is left 2b, up 2c from point P. h. The midpoints (b – a, c) and (a – b, c) are similar but have opposite x-­‐
coordinates so they are not the same point and the diagonals do not bisect each other. R(2b, 2c)
(a-b, c)
T(0, 0)
R(2b, 2c)
M(b, c)
P(2a, 0)
R(2b, 2c)
(d, c)
T(0, 0)
(b-a, c)
P(2a, 0)
A(2d, 2c)
N(a+d, c)
T(0, 0)
A(2a-2b, 2c)
Name:
Date:
A(2d, 2c)
(a+b, c)
P(2a, 0)