The area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3) is:
Answer: B
We can find the area by dividing the quadrilateral into two triangles, e.g., ABC and ADC, and summing their areas. Let A=(-4,-2), B=(-3,-5), C=(3,-2), D=(2,3).
Area(\(\triangle ABC\)) = \(\frac{1}{2} |-4(-5-(-2)) -3(-2-(-2)) + 3(-2-(-5))| = \frac{1}{2} |-4(-3) -3(0) + 3(3)| = \frac{1}{2}|12+9| = 10.5\).
Area(\(\triangle ADC\)) = \(\frac{1}{2} |-4(-2-3) + 3(3-(-2)) + 2(-2-(-2))| = \frac{1}{2} |-4(-5) + 3(5) + 2(0)| = \frac{1}{2}|20+15| = 17.5\).
Total Area = 10.5 + 17.5 = 28 sq units.
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In which quadrant does the point (-3, 5) lie?
Answer: B
The Cartesian coordinate system is divided into four quadrants.
Since the point has a negative x-coordinate (-3) and a positive y-coordinate (5), it lies in the Second Quadrant.
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What is the slope of a line parallel to the x-axis?
Answer: A
A line parallel to the x-axis is a horizontal line. For any two points on a horizontal line, the y-coordinate is constant. The slope formula is \(m = \frac{y_2-y_1}{x_2-x_1}\). Since \(y_2 = y_1\), the numerator is 0.
Therefore, the slope of any horizontal line is 0.
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What is the slope of a line that is perpendicular to the line y = -3x + 4?
Answer: C
The given line is in the form y = mx + c, so its slope (m₁) is -3.
If two lines are perpendicular, the product of their slopes is -1. Let the slope of the perpendicular line be m₂.
m₁ × m₂ = -1
-3 × m₂ = -1
m₂ = \(\frac{-1}{-3} = \frac{1}{3}\).
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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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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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What is the equation of a line that passes through the origin and has a slope of -4?
Answer: A
The slope-intercept form is y = mx + c. A line passing through the origin (0,0) has a y-intercept (c) of 0.
Given the slope m = -4, the equation becomes y = -4x + 0, or simply y = -4x.
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What is the slope of the line given by the equation 5x + 2y = 7?
Answer: B
To find the slope, we rearrange the equation into the slope-intercept form (y = mx + c).
5x + 2y = 7
2y = -5x + 7
y = \(-\frac{5}{2}x + \frac{7}{2}\)
The slope (m) is the coefficient of x, which is -5/2.
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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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What are the coordinates of the midpoint of the line segment joining the points (1, 5) and (7, 3)?
Answer: B
The midpoint formula is \((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\).
Midpoint = \((\frac{1+7}{2}, \frac{5+3}{2})\)
= \((\frac{8}{2}, \frac{8}{2})\)
= (4, 4).
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