The equation of a line with x-intercept 'a' and y-intercept 'b' is given by:
Answer: A
The intercept form of a line is
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
Setting \( y = 0 \) gives \( x = a \), and \( x = 0 \) gives \( y = b \) - which is the check that the intercepts are right.
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Find the value of 'a' for which the points (a, 1), (1, -1), and (11, 4) are collinear.
Answer: B
Correct answer: 5
The two fully known points fix the line. Its slope is
\[m = \frac{4 - (-1)}{11 - 1} = \frac{5}{10} = \frac{1}{2}\]
For (a, 1) to lie on the same line, the slope from it to (1, -1) must also be \(\tfrac{1}{2}\):
\[\frac{-1 - 1}{1 - a} = \frac{1}{2} \quad\Longrightarrow\quad \frac{-2}{1-a} = \frac{1}{2} \quad\Longrightarrow\quad -4 = 1 - a \quad\Longrightarrow\quad a = 5\]
Checking with the third point: the slope from (5, 1) to (11, 4) is \(\tfrac{3}{6} = \tfrac{1}{2}\), as required.
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What is the distance between the points (3, 4) and (-3, -4) ?
Answer: A
Correct answer: 10
\[d = \sqrt{(-3-3)^2 + (-4-4)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\]
Both points lie 5 units from the origin on the same line through it, so the distance is twice that, which is a useful check.
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What is the equation of a line that passes through the origin and has a slope of -4?
Answer: A
A line through the origin has zero intercept, so \( y = mx \).
\[ y = -4x \]
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The points (1, 1), (3, 4), and (5, 1) are the vertices of a/an:
Answer: B
Let's find the lengths of the sides using the distance formula.
Side 1 (between (1,1) and (3,4)): \(\sqrt{(3-1)^2 + (4-1)^2} = \sqrt{2^2 + 3^2} = \sqrt{4+9} = \sqrt{13}\).
Side 2 (between (3,4) and (5,1)): \(\sqrt{(5-3)^2 + (1-4)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}\).
Side 3 (between (1,1) and (5,1)): \(\sqrt{(5-1)^2 + (1-1)^2} = \sqrt{4^2 + 0^2} = 4\).
Since two sides have the same length (\(\sqrt{13}\)), the triangle is isosceles.
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Find the coordinates of the circumcenter of the triangle whose vertices are (8, 6), (8, -2) and (2, -2).
Answer: A
First, notice the vertices. The side from (8, 6) to (8, -2) is a vertical line (x=8). The side from (8, -2) to (2, -2) is a horizontal line (y=-2). This means they are perpendicular, and the triangle is a right-angled triangle with the right angle at (8, -2).
In a right-angled triangle, the circumcenter is the midpoint of the hypotenuse.
The hypotenuse connects the vertices (8, 6) and (2, -2).
Midpoint = \((\frac{8+2}{2}, \frac{6+(-2)}{2}) = (\frac{10}{2}, \frac{4}{2}) = (5, 2)\).
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What is the x-intercept of the line 4x - 5y = 20?
Answer: C
The x-intercept is where \( y = 0 \).
\[ 4x = 20 \quad\Longrightarrow\quad x = 5 \]
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What is the distance of the point (8, -6) from the y-axis?
Answer: C
The distance from the y-axis is the magnitude of the \( x \)-coordinate.
\[ |8| = 8 \]
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Find the value of 'k' if the distance between the points (4, k) and (1, 0) is 5.
Answer: C
\[ \sqrt{(4-1)^{2} + (k-0)^{2}} = 5 \]
\[ 9 + k^{2} = 25 \quad\Longrightarrow\quad k^{2} = 16 \]
\[ k = \pm 4 \]
Both signs satisfy the distance, since squaring loses the sign - the point may lie above or below.
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The reflection of the point (-2, 7) in the y-axis is:
Answer: A
Reflecting in the y-axis flips the sign of \( x \) and leaves \( y \) alone.
\[ (-2,\ 7) \ \rightarrow \ (2,\ 7) \]
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