Find the equation of the line passing through the point (2, 3) with a slope of 4.
Answer: A
Use the point-slope form.
\[ y - 3 = 4(x - 2) \quad\Longrightarrow\quad y = 4x - 8 + 3 \]
\[ y = 4x - 5 \]
Checking: at \( x = 2 \), \( y = 3 \).
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The points A(1, 5), B(2, 3) and C(-2, 11) are:
Answer: A
Correct answer: Collinear
Three points lie on one line exactly when the slopes between them agree.
\[m_{AB} = \frac{3-5}{2-1} = \frac{-2}{1} = -2\]
\[m_{AC} = \frac{11-5}{-2-1} = \frac{6}{-3} = -2\]
Both segments have slope \(-2\) and share the point A, so A, B and C lie on a single straight line and cannot form a triangle of any kind. The line itself is \(y = -2x + 7\), and each of the three points satisfies it.
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The points A(1, 5), B(2, 3) and C(-2, 11) are:
Answer: A
Three points are collinear if the slope between any two pairs of points is the same.
Slope of AB = \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 5}{2 - 1} = \frac{-2}{1} = -2\).
Slope of BC = \(\frac{y_3 - y_2}{x_3 - x_2} = \frac{11 - 3}{-2 - 2} = \frac{8}{-4} = -2\).
Since the slope of AB is equal to the slope of BC, the points A, B, and C lie on the same straight line and are collinear.
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Find the centroid of a triangle with vertices (2, 4), (6, 4), and (2, 0).
Answer: A
The centroid of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by the formula:
Centroid = \((\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})\)
Centroid = \((\frac{2+6+2}{3}, \frac{4+4+0}{3})\)
Centroid = \((\frac{10}{3}, \frac{8}{3})\).
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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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The equation of the x-axis is:
Answer: B
Every point on the x-axis has \( y = 0 \), so that equation describes it.
\[ y = 0 \]
\( x = 0 \) is the y-axis - the pair most easily transposed.
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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
Correct answer: \(\sqrt{80}\)
The radius is the distance from the centre to any point on the circle, so apply the distance formula to (3, 2) and (-5, 6).
\[r = \sqrt{(-5-3)^2 + (6-2)^2} = \sqrt{(-8)^2 + 4^2} = \sqrt{64+16} = \sqrt{80}\]
This simplifies to \(4\sqrt{5} \approx 8.94\), which is why neither 8 nor 10 is correct - the radius falls between them.
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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
\[ d = \sqrt{x^{2} + y^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
The sign of the coordinate disappears in the squaring.
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