Question

Part 1

Answer 1

Answer:

1) c. AD || CE

2) solve them both first:

3y + 1 = 6x + 4                2y + 1 = x - 9

3y = 6x + 3                     2y = x - 10

y = 2x + 1 .                      y = 1/2x - 5

Parallel lines should have the same slope and they are definitely not the same line, so both A and C are eliminated

Perpendicular slopes should be negative reciprocals of each other. If they were negative reciprocals one slope would be either -2 or -1/2, which means that they're not perpendicular either.

The answer is D, neither parallel nor perpendicular.

3) to find the midpoint of a line, add the x coordinates and divide by 2 and add the y coordinates and divide by 2.

$(\frac{-6+1}{2},\frac{1+8}{2} )\\\\=(\frac{-5}{2},\frac{9}{2} )\\\\=(-2\frac{1}{2} ,4\frac{1}{2} )$

The answer is D.

4) d. GB || HC

5) d. 345.6

$A=2\pi rh\\\\A=2\pi (5)(11)\\\\A=2\pi (55)$

A ≈ 345.58 ≈ 345.6

6) first change the other equation (4x + 2y = 10) to slope-intercept form:

2y = -4x + 10

y = -2x + 5

because the line must be parallel, it has to have a slope = -2

That automatically eliminates all answers except for D, so you don't even have to do the work to make the other equation.

7) best bet is a. EH and BC are coplanar, but I'm really not sure

8) d. DB and HF

9) first change the other equation (4x - 6y = 15) to slope-intercept form:

-6y = -4x + 15

y = 2/3x - 5/2

slope of the perpendicular line should be -3/2

y = -3/2x + a

9 = -3/2(6) + a

9 = -9 + a

a = 18

y = -3/2x + 18

which is the same as y - 9 = -3/2(x - 6), which is A

Answer 2

Answer:

Answer:

1) c. AD || CE

2) solve them both first:

3y + 1 = 6x + 4 2y + 1 = x - 9

3y = 6x + 3 2y = x - 10

y = 2x + 1 . y = 1/2x - 5

Parallel lines should have the same slope and they are definitely not the same line, so both A and C are eliminated

Perpendicular slopes should be negative reciprocals of each other. If they were negative reciprocals one slope would be either -2 or -1/2, which means that they're not perpendicular either.

The answer is D, neither parallel nor perpendicular.

3) to find the midpoint of a line, add the x coordinates and divide by 2 and add the y coordinates and divide by 2.

\begin{gathered}(\frac{-6+1}{2},\frac{1+8}{2} )\\\\=(\frac{-5}{2},\frac{9}{2} )\\\\=(-2\frac{1}{2} ,4\frac{1}{2} )\end{gathered}

(

2

−6+1

,

2

1+8

)

=(

2

−5

,

2

9

)

=(−2

2

1

,4

2

1

)

The answer is D.

4) d. GB || HC

5) d. 345.6

\begin{gathered}A=2\pi rh\\\\A=2\pi (5)(11)\\\\A=2\pi (55)\end{gathered}

A=2πrh

A=2π(5)(11)

A=2π(55)

A ≈ 345.58 ≈ 345.6

6) first change the other equation (4x + 2y = 10) to slope-intercept form:

2y = -4x + 10

y = -2x + 5

because the line must be parallel, it has to have a slope = -2

That automatically eliminates all answers except for D, so you don't even have to do the work to make the other equation.

7) best bet is a. EH and BC are coplanar, but I'm really not sure

8) d. DB and HF

9) first change the other equation (4x - 6y = 15) to slope-intercept form:

-6y = -4x + 15

y = 2/3x - 5/2

slope of the perpendicular line should be -3/2

y = -3/2x + a

9 = -3/2(6) + a

9 = -9 + a

a = 18

y = -3/2x + 18

which is the same as y - 9 = -3/2(x - 6), which is A