Math 7A
EXAM 2
Find the distance d(P1 , P2 ) between the points P1 and P2 .
1) P1 = (-3 12, 1); P2 = (-8 27, -11)
-3 12 - -8 27 2 + 1 - - 11 2
-6 3 + 24 3 2 + 12 2
d=
d=
d=
d=
d=
d=
d=6
18 3 2 + 144
324 3 + 144
972+ 144
1116
31
Solve the problem.
2) If (5, -2) is the endpoint of a line segment, and (9, 3) is its midpoint, find the other endpoint.
9=
x+ 5
2
18 = x + 5
13 = x
3=
y + (- 2)
2
6=y-2
8 =y
13, 8
Find and identify the intercepts for the given equation.
x2 - 16
3) y =
4x4
x2 - 16
4x4
y=
0 2 - 16
404
0 = x 2 - 16
y=
- 16
0
16 = x2
±4 = x
x-intercepts ±4, 0
undefined, therefore no y-intercept
0=
1
FALL 2011
Find and identify the slope and intercepts of the line.
6
9
x = 15y 4)
20
2
6
9
x - 15y = 20
2
m=m=
6/20
- 15
6 -1
20 15
m=-
1
50
-9/2
6/20
-
-9/2
- 15
9 20
2 6
-
9 -1
2 15
- 15
3
10
x-intercept - 15, 0
y-intercept 0,
3
10
Determine whether the graph of the equation is symmetric with respect to the x-axis, the y-axis, and/or the origin.
9x
5) y =
x2 + 81
x-axis test
9x
-y=
2
x + 81
y=no
9x
2
x + 81
y-axis test
9( - x)
y=
( - x)2 + 81
y=-
9x
2
x + 81
no
origin test
9( - x)
-y=
( - x)2 + 81
-y=y=
Plot six points so that it shows the given type of symmetry.
6) Symmetric with respect to the x-axis
2
9x
2
x + 81
9x
x2 + 81
yes
Find an equation for the line, in the indicated form, with the given properties.
3
7 12
); general form CORRECTION: standard form
7) Containing the points (- , -2) and ( ,
8
10 5
-2m=
12
5
3
7
- 8 10
y - -2 =
·
40 - 80 - 96 - 176 176
=
=
=
40 - 15 - 28
43
- 43
176
3
x- 43
8
43 y + 2 = 176 x +
3
8
43y + 86 = 176x + 66
20 = 176x - 43y
Find an equation for the line with the given properties. Identify the slope of each line; clearly.
8) Perpendicular to the line 3y + 15 = -9; containing the point (7, 5)
3y + 15 = -9
3y = - 24
y=- 8
m=0
m = undefined
x=7
Find the standard form of the equation of the the circle, then find the center (h, k) and radius r.
9) 4x2 + 4y2 - 12x + 16y - 5 = 0
4(x2 - 3x )+ 4(y2 + 4y ) = 5
9
4 x2 - 3x +
+ 4 y2 + 4y + 4 = 5 + 9 + 16
4
3 2
+ 4 x + 2 2 = 30
4 x2
3 2
30
x+ x+22=
2
4
2
3
15
+ x+22=
x2
2
C
3
,- 2
2
I used
r=
30
=
4
30
2
30
15
instead of
because 4 is a perfect square and I wouldn't need to rationalize
4
2
Find the standard form of the equation of the the circle, with the given properties.
10) Center at the point (5, 7); tangent to y-axis
x - 5 2 + x - 7 2 = 52
x - 5 2 + x - 7 2 = 25
3
Solve the problem by first writing a variation equation and using it to determine the final answer. (10 points)
11) Body-mass index, or BMI, takes both weight and height into account when assessing whether an individual is
underweight or overweight. BMI varies directly as one's weight, in pounds, and inversely as the square of one's
height, in inches. In adults, normal values for the BMI are between 20 and 25. A person who weighs 171 pounds
and is 72 inches tall has a BMI of 23.19. What is the BMI, to the nearest tenth, for a person who weighs 136
pounds 15 ounces and who is 5feet 5 inches tall? Would this person be considered to have a normal BMI?
B=
kW
H2
23.19 =
k=
k 171
72 2
B=
23.19 72 2
171
B=
B=
B=
yes, this person would be considered to have normal BMI.
4
B
23.19 72 2
15
136
171
16
5(12) + 5 2
23.19 5184
171
2191
16
60 + 5 2
23.19 5184
171
2191
16
65 2
23.19 5184
171
22.8
2191
1
16
4225