Are my answers correct?
Identify which lines are parallel.
1. y = 6; y = 6x + 5; y = 6x - 7; y = -8
A: y = 6; y = -8; y = 6x +5; y = 6x - 7 (all)
2. y = 3/4x - 1; y = -2x; y - 3 = 3/4(x - 5); y - 4 = -2(x + 2)
A: y = 3/4x - 1; y- 3 = 3/4(x -5)
Identify which lines are perpendicular.
3. y = 2/3x - 4; y = -3/2x + 2; y = -1; x = 3
A: y = 2/3x - 4; y = -3/2x + 2; y = -1; x = 3 (all)
4. y = -3/7x - 4; y - 4 = -7(x + 2); y - 1 = 1/7(x - 4); y - 7 = 7/3(x - 3)
A: y = -3/7x - 4; y - 4 = -7(x + 2); y - 1 = 1/7(x - 4); y - 7 = 7/3(x - 3) (all)
5. Show that PQRS is a rectangle.
(P: (1,4); Q: (2,6); R: (8,3); S: (7,1)
A: PQ is perpendicular to QR because 2(-1/2) = -1. RS is perpendicular to PS because 2(-1/2) = -1. Therefore. RQRS is a rectangle because it contains a right angle.
1. y=6 and y=-8 are parallel, they are both horizontal
y=6x + 5 and y = 6x-7 are parallel, they both have a slope of 6
2. the first and third are parallel, you had that
but also the 2nd and last line are parallel, both have a slope of -2
3. the 1st and 2nd lines are perpendicular
the last two , y = -1, and x=3 are perpendicualar.
The y = -1 is horizontal, x=3 is vertical
4. 1st and last, as well as the 2nd and third
5.You have shown that opposite angles are 90° each, but I can draw a quad where opposite angles are each 90° , one other is 80, and the third is 100
so you have to show one more vertex to have a 90°
(Then 3 90's would leave 90 for the fourth angle)
Can someone please tell me whether these answers are correct or not?
5. I don't quite understand what you said. Have I answered correctly?
#5.
no, it is not a rectangle.
angle P = 90°
angle S = 90°
but since slope PQ = 2 and slope of QR = -3/5
we don't have a right angle at Q
no need to go further.
But I thought PQ = -1/2.
Oh. It appears that I have made a mistake.
Correction: But I thought that QR = -1/2.
To check if lines are parallel, you need to compare their slopes. If two lines have the same slope, they are parallel.
In question 1, the lines are:
- y = 6
- y = 6x + 5
- y = 6x - 7
- y = -8
To identify which lines are parallel, we need to compare their slopes. The slopes of all lines are:
- y = 6: This line has a slope of 0, since it is a horizontal line.
- y = 6x + 5: This line has a slope of 6.
- y = 6x - 7: This line also has a slope of 6.
- y = -8: This line has a slope of 0, since it is a horizontal line.
So, the lines with slopes of 6 (y = 6x + 5 and y = 6x - 7) are parallel to each other. The lines with slopes of 0 (y = 6 and y = -8) are also parallel to each other.
Therefore, the answer is: y = 6; y = -8; y = 6x + 5; y = 6x - 7 (all lines) are parallel.
Now let's move on to question 2:
The lines are:
- y = 3/4x - 1
- y = -2x
- y - 3 = 3/4(x - 5)
- y - 4 = -2(x + 2)
To identify which lines are parallel, we need to compare their slopes. The slopes of all lines are:
- y = 3/4x - 1: This line has a slope of 3/4.
- y = -2x: This line has a slope of -2.
- y - 3 = 3/4(x - 5): This line has a slope of 3/4.
- y - 4 = -2(x + 2): This line has a slope of -2.
So, the lines with slopes of 3/4 (y = 3/4x - 1 and y - 3 = 3/4(x - 5)) are parallel to each other.
Therefore, the answer is: y = 3/4x - 1; y - 3 = 3/4(x - 5) are parallel.
Moving on to identifying perpendicular lines in question 3 and 4:
To check if lines are perpendicular, you need to compare their slopes. If the product of the slopes is -1, the lines are perpendicular.
In question 3, the lines are:
- y = 2/3x - 4
- y = -3/2x + 2
- y = -1
- x = 3
To identify which lines are perpendicular, we need to compare their slopes. The slopes of all lines are:
- y = 2/3x - 4: This line has a slope of 2/3.
- y = -3/2x + 2: This line has a slope of -3/2.
- y = -1: This line is horizontal and has a slope of 0.
- x = 3: This line is vertical and has an undefined slope.
The product of the slopes of y = 2/3x - 4 and y = -3/2x + 2 is (-2/3) * (3/2) = -1, which is -1.
Therefore, the lines y = 2/3x - 4 and y = -3/2x + 2 are perpendicular.
In question 4, the lines are:
- y = -3/7x - 4
- y - 4 = -7(x + 2)
- y - 1 = 1/7(x - 4)
- y - 7 = 7/3(x - 3)
To identify which lines are perpendicular, we need to compare their slopes. The slopes of all lines are:
- y = -3/7x - 4: This line has a slope of -3/7.
- y - 4 = -7(x + 2): This line has a slope of -7.
- y - 1 = 1/7(x - 4): This line has a slope of 1/7.
- y - 7 = 7/3(x - 3): This line has a slope of 7/3.
The product of the slopes of y = -3/7x - 4 and y - 4 = -7(x + 2) is (-3/7) * (-7) = 1, which is not -1.
Therefore, none of the lines in question 4 are perpendicular to each other.
Moving on to question 5:
To show that PQRS is a rectangle, we need to check if the opposite sides are parallel and if the diagonals are congruent.
The points given are:
- P: (1,4)
- Q: (2,6)
- R: (8,3)
- S: (7,1)
To check if PQ is perpendicular to QR, we need to compare their slopes:
- PQ: (1,4) to (2,6) has a slope of (6-4)/(2-1) = 2/1 = 2
- QR: (2,6) to (8,3) has a slope of (3-6)/(8-2) = -3/6 = -1/2
The product of the slopes is 2 * (-1/2) = -1, which means PQ is perpendicular to QR.
To check if RS is perpendicular to PS, we need to compare their slopes:
- RS: (8,3) to (7,1) has a slope of (1-3)/(7-8) = -2/-1 = 2
- PS: (1,4) to (7,1) has a slope of (1-4)/(7-1) = -3/6 = -1/2
The product of the slopes is 2 * (-1/2) = -1, which means RS is perpendicular to PS.
Since both pairs of opposite sides are perpendicular, PQRS is a rectangle.
Therefore, the statement that PQRS is a rectangle is true because it contains right angles.