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 distance between the points (3, 2) and (6, 6)?
Answer: C
The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).
Distance = \(\sqrt{(6-3)^2 + (6-2)^2}\)
= \(\sqrt{3^2 + 4^2}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25} = 5\) units.
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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 averages the coordinates.
\[ \left(\frac{1+7}{2},\ \frac{5+3}{2}\right) = (4,\ 4) \]
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What is the slope of the line passing through the points (2, 3) and (4, 7)?
Answer: B
\[ m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{7-3}{4-2} = 2 \]
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What is the equation of a line with a slope of 3 and a y-intercept of 5?
Answer: B
The slope-intercept form is \( y = mx + c \).
\[ y = 3x + 5 \]
The slope multiplies \( x \) and the intercept stands alone - swapping them gives the first option.
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The point (0, -4) lies on:
Answer: B
A point with \( x = 0 \) lies on the y-axis, whatever its \( y \)-coordinate.
The origin is the single point \( (0,0) \), and the first quadrant needs both coordinates positive.
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Find the coordinates of the point which divides the line segment joining (-1, 7) and (4, -3) in the ratio 2:3.
Answer: A
Using the section formula, the coordinates (x, y) are given by:
x = \(\frac{m_1x_2 + m_2x_1}{m_1+m_2}\) and y = \(\frac{m_1y_2 + m_2y_1}{m_1+m_2}\)
Here, \((x_1, y_1) = (-1, 7)\), \((x_2, y_2) = (4, -3)\), and \(m_1:m_2 = 2:3\).
x = \(\frac{2(4) + 3(-1)}{2+3} = \frac{8-3}{5} = \frac{5}{5} = 1\)
y = \(\frac{2(-3) + 3(7)}{2+3} = \frac{-6+21}{5} = \frac{15}{5} = 3\)
The coordinates are (1, 3).
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What is the area of the triangle with vertices (1, -1), (-4, 6) and (-3, -5)?
Answer: B
The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by the formula:
Area = \(\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|\)
Area = \(\frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|\)
Area = \(\frac{1}{2} |1(11) + (-4)(-4) + (-3)(-7)|\)
Area = \(\frac{1}{2} |11 + 16 + 21| = \frac{1}{2} |48| = 24\) sq units.
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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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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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