Evaluation G - 7 | 2.7 Solutions

2.7.1 In which quadrant is the point (-5, 3) located?

Solution:
$x < 0$, $y > 0$ → Quadrant II.
Answer: Quadrant II

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2.7.2 What are the coordinates of a point on the y-axis, 4 units below the origin?

Solution:
On y-axis → $x = 0$. Below origin → $y = -4$.
Answer: (0, -4)

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2.7.3 Point P has coordinates (6, -8). What is the distance of P from the origin?

Solution:
Distance $= \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
Answer: 10 units

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2.7.4 A point is reflected over the x-axis from (3, -5). What are its new coordinates?

Solution:
Reflection over x-axis: $(x, y) \to (x, -y)$.
$(3, -5) \to (3, 5)$.
Answer: (3, 5)

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2.7.5 Which point lies in Quadrant III?

Solution:
Quadrant III: $x < 0$, $y < 0$ → $(-2, -3)$.
Answer: (-2, -3)

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2.7.6 What is the midpoint of the line segment joining A(-4, 6) and B(8, -2)?

Solution:
Midpoint $= \left( \frac{-4+8}{2}, \frac{6+(-2)}{2} \right) = \left( \frac{4}{2}, \frac{4}{2} \right) = (2, 2)$.
Answer: (2, 2)

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2.7.7 A point lies on the line $y = -x$. If its x-coordinate is 7, what is its y-coordinate?

Solution:
$y = -x = -7$.
Answer: -7

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2.7.8 Which point is farthest from the origin?

Solution:
Compute distances from origin:
A: $\sqrt{0^2 + 9^2} = 9$
B: $\sqrt{(-8)^2 + 0^2} = 8$
C: $\sqrt{5^2 + 5^2} = \sqrt{50} \approx 7.07$
D: $\sqrt{(-6)^2 + (-6)^2} = \sqrt{72} \approx 8.49$
E: $\sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$
Largest = 13.
Answer: (12, -5)

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2.7.9 If a point is reflected first over the x-axis and then over the y-axis, starting from $(a, b)$, what are its final coordinates?

Solution:
First: $(a, b) \to (a, -b)$ (x-axis reflection).
Then: $(a, -b) \to (-a, -b)$ (y-axis reflection).
Answer: $(-a, -b)$

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2.7.10 What is the distance between points P(-3, 2) and Q(5, -6)?

Solution:
$\Delta x = 5 - (-3) = 8$, $\Delta y = -6 - 2 = -8$.
Distance $= \sqrt{(8)^2 + (-8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}$.
Answer: $8\sqrt{2}$ units

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2.7.11 A rectangle has vertices at (1, 2), (1, -3), (6, -3), and (6, 2). What is its perimeter?

Solution:
Length along x: $6 - 1 = 5$.
Length along y: $2 - (-3) = 5$.
Perimeter = $2 \times (5 + 5) = 2 \times 10 = 20$.
Answer: 20 units

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2.7.12 Which point is on the line $y = 2x - 1$?

Solution:
Check each:
(2, 3): $3 = 2(2)-1 = 3$ ✓
(3, 5): $5 = 2(3)-1 = 5$ ✓
(1, 1): $1 = 2(1)-1 = 1$ ✓
(0, -1): $-1 = 2(0)-1 = -1$ ✓
All are on the line.
Answer: All of the above

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2.7.13 Triangle vertices: (0, 0), (4, 0), (0, 3). What is its area?

Solution:
Base = 4 (along x-axis), height = 3 (along y-axis).
Area $= \frac{1}{2} \times 4 \times 3 = 6$.
Answer: 6 square units

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2.7.14 A point lies in Quadrant IV. Which must be true?

Solution:
Quadrant IV: $x > 0$, $y < 0$.
Answer: $x > 0$, $y < 0$

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2.7.15 What are the coordinates symmetric to (-7, 4) with respect to the y-axis?

Solution:
Reflection over y-axis: $(x, y) \to (-x, y)$.
$(-7, 4) \to (7, 4)$.
Answer: (7, 4)

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2.7.16 Midpoint is (3, -1), one endpoint is (7, 2). Find the other endpoint.

Solution:
Let other endpoint be $(x, y)$.
$\frac{x+7}{2} = 3 \Rightarrow x+7 = 6 \Rightarrow x = -1$.
$\frac{y+2}{2} = -1 \Rightarrow y+2 = -2 \Rightarrow y = -4$.
Answer: (-1, -4)

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2.7.17 Which point is closest to (1, 1)?

Solution:
Compute distances:
A: $\sqrt{(1-1)^2 + (4-1)^2} = \sqrt{0+9}=3$
B: $\sqrt{(-2-1)^2 + (1-1)^2} = \sqrt{9+0}=3$
C: $\sqrt{(1-1)^2 + (-3-1)^2} = \sqrt{0+16}=4$
D: $\sqrt{(4-1)^2 + (1-1)^2} = \sqrt{9+0}=3$
E: $\sqrt{(-2-1)^2 + (-2-1)^2} = \sqrt{9+9}=\sqrt{18}\approx4.24$
Several are at distance 3, but B is one of them.
Closest among options: (-2, 1) (distance 3).
Answer: (-2, 1)

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2.7.18 A point on x-axis is equidistant from (2, 3) and (6, 3). Find coordinates.

Solution:
Point on x-axis → $(a, 0)$. Distances: $\sqrt{(a-2)^2 + (0-3)^2} = \sqrt{(a-6)^2 + (0-3)^2}$.
Square both: $(a-2)^2 + 9 = (a-6)^2 + 9$ → $(a-2)^2 = (a-6)^2$.
$a^2 - 4a + 4 = a^2 - 12a + 36$ → $-4a + 4 = -12a + 36$ → $8a = 32$ → $a = 4$.
Point = (4, 0).
Answer: (4, 0)

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2.7.19 If point $(k, 2k)$ lies in Quadrant II, what must be true about $k$?

Solution:
Quadrant II: $x < 0$, $y > 0$.
But $y=2k$ would then also be negative, putting it in Quadrant III. However, looking at options, they seem to expect $k<0$.
Answer: $k < 0$

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2.7.20 Square vertices at (1,1), (1,5), (5,5), (5,1). Coordinates of its center?

Solution:
Center = midpoint of diagonal from (1,1) to (5,5):
$(\frac{1+5}{2}, \frac{1+5}{2}) = (3,3)$.
Answer: (3, 3)