The distance of the point P(4, -3) from the origin is:
Answer: B
The origin has coordinates (0, 0). We use the distance formula.
Distance = \(\sqrt{(4-0)^2 + (-3-0)^2}\)
= \(\sqrt{4^2 + (-3)^2}\)
= \(\sqrt{16 + 9}\)
= \(\sqrt{25} = 5\) units.
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What is the equation of the line that is parallel to y = 2x + 3 and passes through the point (1, 7)?
Answer: A
Parallel lines have the same slope. The slope of the given line is m=2.
The new line also has a slope of 2 and passes through (1, 7).
Using the point-slope form y - y₁ = m(x - x₁):
y - 7 = 2(x - 1)
y - 7 = 2x - 2
y = 2x + 5
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Three vertices of a parallelogram taken in order are (-1, 0), (3, 1) and (2, 2). The coordinates of the fourth vertex are:
Answer: A
In a parallelogram, the diagonals bisect each other. This means the midpoint of one diagonal is the same as the midpoint of the other diagonal.
Let the vertices be A(-1, 0), B(3, 1), C(2, 2), and D(x, y).
Midpoint of diagonal AC = \((\frac{-1+2}{2}, \frac{0+2}{2}) = (1/2, 1)\).
Midpoint of diagonal BD = \((\frac{3+x}{2}, \frac{1+y}{2})\).
Equating the midpoints: \(\frac{3+x}{2} = \frac{1}{2} \Rightarrow 3+x=1 \Rightarrow x=-2\).
\(\frac{1+y}{2} = 1 \Rightarrow 1+y=2 \Rightarrow y=1\).
The fourth vertex is (-2, 1).
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What is the area of a rhombus whose vertices are (3,0), (4,5), (-1,4), and (-2,-1)?
Answer: B
The area of a rhombus can be calculated as half the product of its diagonals.
Let the vertices be A(3,0), B(4,5), C(-1,4), D(-2,-1).
Length of diagonal AC = \(\sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}\).
Length of diagonal BD = \(\sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}\).
Area = \(\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24\) sq units.
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Find the y-intercept of the line passing through the points (2, 5) and (4, 9).
Answer: A
First, find the slope (m) of the line.
m = \(\frac{9-5}{4-2} = \frac{4}{2} = 2\).
Now use the point-slope form with one of the points, say (2, 5):
y - 5 = 2(x - 2)
y - 5 = 2x - 4
y = 2x + 1.
This is in the slope-intercept form y = mx + c. The y-intercept (c) is 1.
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The line \(2x + 3y = 6\) cuts the y-axis at which point?
Answer: D
A line cuts the y-axis at a point where the x-coordinate is 0.
Set x = 0 in the equation:
2(0) + 3y = 6
3y = 6
y = 2.
The point is (0, 2).
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The distance of the point (5, 12) from the x-axis is:
Answer: B
The distance of any point (x, y) from the x-axis is the absolute value of its y-coordinate.
For the point (5, 12), the distance from the x-axis is |12| = 12 units.
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Find the value of k if the line \(kx + 3y - 1 = 0\) is parallel to the line \(2x + y + 5 = 0\).
Answer: C
For lines to be parallel, their slopes must be equal.
Slope of the first line \(kx+3y-1=0 \Rightarrow 3y = -kx+1 \Rightarrow y = (-k/3)x+1/3\). Slope m₁ = -k/3.
Slope of the second line \(2x+y+5=0 \Rightarrow y = -2x-5\). Slope m₂ = -2.
Set m₁ = m₂:
-k/3 = -2
k = 6.
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The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x + 3. Find the third vertex.
Answer: D
Let the third vertex be (x, y). Since it lies on y = x+3, its coordinates are (x, x+3).
Using the area formula: Area = \(\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|\)
5 = \(\frac{1}{2} |2(-2-(x+3)) + 3((x+3)-1) + x(1-(-2))|\)
10 = |2(-x-5) + 3(x+2) + x(3)|
10 = |-2x - 10 + 3x + 6 + 3x| = |4x - 4|
This gives two possibilities:
1) 4x - 4 = 10 => 4x = 14 => x = 14/4 = 7/2. Then y = 7/2 + 3 = 13/2. Vertex is (7/2, 13/2).
2) 4x - 4 = -10 => 4x = -6 => x = -6/4 = -3/2. Then y = -3/2 + 3 = 3/2. Vertex is (-3/2, 3/2).
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The reflection of the point (3, -5) in the x-axis is:
Answer: C
When a point (x, y) is reflected in the x-axis, its x-coordinate remains the same, and its y-coordinate changes sign.
The reflection of (3, -5) in the x-axis is (3, 5).
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