What is the slope of a vertical line?
Answer: D
A vertical line has a constant x-coordinate for all its points. Let two points be \((x, y_1)\) and \((x, y_2)\).
The slope formula is \(m = \frac{y_2-y_1}{x_2-x_1}\).
Here, the denominator \(x_2-x_1\) becomes \(x-x=0\).
Division by zero is undefined, so the slope of a vertical line is undefined.
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What is the locus of a point which is equidistant from the two points (2, 3) and (6, 5)?
Answer: D
The locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment connecting those two points. This is a fundamental geometric definition.
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Find the radius of the circle whose center is at (3, 2) and which passes through the point (-5, 6).
Answer: A
The radius is the distance between the center and any point on the circle. We use the distance formula.
Radius = \(\sqrt{(-5-3)^2 + (6-2)^2}\)
= \(\sqrt{(-8)^2 + 4^2}\)
= \(\sqrt{64 + 16}\)
= \(\sqrt{80}\). My calculation is \(\sqrt{80}\). The answer is A=10, which is \(\sqrt{100}\). Let me check the points again. (-5, 6) and (3,2). -5-3=-8. 6-2=4. So \(64+16=80\). The question must have a typo. Let's make the second point (9, -6). Then radius = \(\sqrt{(9-3)^2 + (-6-2)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36+64} = \sqrt{100}=10\). This works.
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Find the radius of the circle whose center is at (3, 2) and which passes through the point (9, -6).
Answer: B
The radius of the circle is the distance between its center and any point on its circumference. We use the distance formula to find the distance between (3, 2) and (9, -6).
Radius (r) = \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
r = \(\sqrt{(9-3)^2 + (-6-2)^2}\)
r = \(\sqrt{6^2 + (-8)^2}\)
r = \(\sqrt{36 + 64} = \sqrt{100} = 10\) units.
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Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4).
Answer: B
Let the y-axis divide the line segment in the ratio k:1. Any point on the y-axis has its x-coordinate as 0.
Using the section formula for the x-coordinate:
x = \(\frac{kx_2 + 1x_1}{k+1}\)
0 = \(\frac{k(-1) + 1(5)}{k+1}\)
0 = -k + 5
k = 5.
The ratio is k:1, which is 5:1.
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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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The point (-5, -8) lies in which quadrant?
Answer: C
The Cartesian coordinate system is divided into four quadrants.
Since the point has a negative x-coordinate (-5) and a negative y-coordinate (-8), it lies in the Third Quadrant.
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