Writing Line Equations - Accuplacer Qas

Line L passes through the origin and the point $(a, b)$ , where $a \neq 0$ and $b \neq 0$ . Line M is perpendicular to line L and also passes through the origin. Which of the following is an equation for line M?
Answer: $y = -\frac{a}{b}x$ The slope of line L, which passes through $(0, 0)$ and $(a, b)$ , is $m_L = \frac{b - 0}{a - 0} = \frac{b}{a}$ . Line M is perpendicular to line L, so its slope $m_M$ must be the negative reciprocal of $m_L$ . Therefore, $m_M = -\frac{a}{b}$ . Since line M also passes through the origin, its y-intercept is 0. The equation for line M is $y = m_M x + b$ , which becomes $y = -\frac{a}{b}x + 0$ , or $y = -\frac{a}{b}x$ .
A line segment has endpoints at $A(-2, 1)$ and $B(4, 5)$ . What is the equation of the perpendicular bisector of the segment AB?
Answer: $y = -\frac{3}{2}x + \frac{9}{2}$ A perpendicular bisector passes through the midpoint of the segment and has a slope that is the negative reciprocal of the segment's slope. First, find the midpoint: $(\frac{-2+4}{2}, \frac{1+5}{2}) = (1, 3)$ . Next, find the slope of segment AB: $m_{AB} = \frac{5 - 1}{4 - (-2)} = \frac{4}{6} = \frac{2}{3}$ . The slope of the perpendicular bisector is the negative reciprocal, $m_{\perp} = -\frac{3}{2}$ . Finally, use the point-slope form with the midpoint $(1, 3)$ and perpendicular slope: $y - 3 = -\frac{3}{2}(x - 1)$ . Simplifying gives $y - 3 = -\frac{3}{2}x + \frac{3}{2}$ , so $y = -\frac{3}{2}x + \frac{3}{2} + 3 = -\frac{3}{2}x + \frac{9}{2}$ .
A line has an x-intercept of -4 and a y-intercept of 6. Which of the following is an equation of the line?
Answer: $y = \frac{3}{2}x + 6$ The x-intercept of -4 corresponds to the point $(-4, 0)$ . The y-intercept of 6 corresponds to the point $(0, 6)$ . The y-intercept value directly gives $b=6$ in the slope-intercept form $y = mx + b$ . To find the slope $m$ , use the two points: $m = \frac{6 - 0}{0 - (-4)} = \frac{6}{4} = \frac{3}{2}$ . Combining the slope and y-intercept gives the equation $y = \frac{3}{2}x + 6$ .
A line passes through the point $(2, -3)$ and is perpendicular to the line $3x - 4y = 12$ . What is the equation of this line in slope-intercept form?
Answer: $y = -\frac{4}{3}x + \frac{1}{3}$ First, find the slope of the given line by rewriting in slope-intercept form: $3x - 4y = 12$ becomes $y = \frac{3}{4}x - 3$ , so the slope is $\frac{3}{4}$ . A perpendicular line has slope $-\frac{4}{3}$ . Using point-slope form with $(2, -3)$ : $y - (-3) = -\frac{4}{3}(x - 2)$ , which simplifies to $y = -\frac{4}{3}x + \frac{8}{3} - 3 = -\frac{4}{3}x + \frac{1}{3}$ . Choice B uses the wrong y-intercept (sign error). Choice C uses the slope of the original line instead of perpendicular slope. Choice D uses the negative reciprocal incorrectly.
The line $2x - 3y + 6 = 0$ is reflected across the x-axis. What is the equation of the reflected line?
Answer: $2x + 3y + 6 = 0$ When a line is reflected across the x-axis, each point $(x, y)$ becomes $(x, -y)$ . The original line is $2x - 3y + 6 = 0$ . To find the reflected line, replace $y$ with $-y$ : $2x - 3(-y) + 6 = 0$ , which simplifies to $2x + 3y + 6 = 0$ . Choice B changes the constant term incorrectly. Choice C changes the x-coefficient unnecessarily. Choice D changes the constant term sign incorrectly.
A line passes through the points $(0, -4)$ and $(6, -1)$ . What is the equation of the line?
Answer: $y = \frac{1}{2}x - 4$ The point $(0, -4)$ indicates that the y-intercept $b$ is -4. To find the slope $m$ , use the two points: $m = \frac{-1 - (-4)}{6 - 0} = \frac{-1 + 4}{6} = \frac{3}{6} = \frac{1}{2}$ . Using the slope-intercept form $y = mx + b$ , the equation is $y = \frac{1}{2}x - 4$ .
A new Plan C is created. It has a slope (per-minute cost) equal to the average of the slopes for Plans A and B, and it passes through the point where the costs of Plan A and Plan B are equal. What is the equation for the cost, $C$ , of Plan C as a function of minutes, $m$ ?
Answer: $C = 0.20m + 10$ First, write the cost equations for Plan A and Plan B: $C_A = 0.10m + 20$ and $C_B = 0.30m$ . Find their intersection point by setting the costs equal: $0.10m + 20 = 0.30m \implies 20 = 0.20m \implies m = 100$ . The cost at this point is $C = 0.30(100) = 30$ . So, the intersection point is $(100, 30)$ . The slope for Plan C is the average of the slopes of A and B: $m_C = \frac{0.10 + 0.30}{2} = \frac{0.40}{2} = 0.20$ . Now, use the point-slope form with the point $(100, 30)$ and slope 0.20: $C - 30 = 0.20(m - 100)$ . Distribute the slope: $C - 30 = 0.20m - 20$ . Add 30 to both sides to get $C = 0.20m + 10$ .
A line has slope $\frac{2}{3}$ and passes through the midpoint of the segment connecting $(-2, 4)$ and $(6, -2)$ . What is the equation of this line in point-slope form?
Answer: $y - 1 = \frac{2}{3}(x - 2)$ The midpoint of the segment from $(-2, 4)$ to $(6, -2)$ is $\left(\frac{-2+6}{2}, \frac{4+(-2)}{2}\right) = (2, 1)$ . Using the point-slope form with slope $\frac{2}{3}$ and point $(2, 1)$ : $y - 1 = \frac{2}{3}(x - 2)$ . Choice B has the wrong sign for the y-coordinate. Choice C has the wrong sign for the x-coordinate. Choice D reverses the coordinates of the midpoint.
A line passes through the points $(2, 5)$ and $(-4, 1)$ . What is the equation of the line?
Answer: $y = \frac{2}{3}x + \frac{11}{3}$ First, calculate the slope $m$ using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ . So, $m = \frac{1 - 5}{-4 - 2} = \frac{-4}{-6} = \frac{2}{3}$ . Next, use the point-slope form, $y - y_1 = m(x - x_1)$ , with the point $(2, 5)$ . This gives $y - 5 = \frac{2}{3}(x - 2)$ . Distributing the slope gives $y - 5 = \frac{2}{3}x - \frac{4}{3}$ . Finally, add 5 (or $\frac{15}{3}$ ) to both sides to isolate $y$ : $y = \frac{2}{3}x - \frac{4}{3} + \frac{15}{3}$ , which simplifies to $y = \frac{2}{3}x + \frac{11}{3}$ .
Which of the following is an equation of the line that passes through the point $(-1, 3)$ and is parallel to the line with the equation $2x + 3y = 6$ ?
Answer: $y = -\frac{2}{3}x + \frac{7}{3}$ First, find the slope of the given line by converting its equation to slope-intercept form ( $y = mx + b$ ). $2x + 3y = 6 \implies 3y = -2x + 6 \implies y = -\frac{2}{3}x + 2$ . The slope is $-\frac{2}{3}$ . A parallel line has the same slope. Now, use the point-slope form $y - y_1 = m(x - x_1)$ with the point $(-1, 3)$ and slope $-\frac{2}{3}$ : $y - 3 = -\frac{2}{3}(x - (-1))$ . This simplifies to $y - 3 = -\frac{2}{3}(x + 1)$ , then $y - 3 = -\frac{2}{3}x - \frac{2}{3}$ . Adding 3 (or $\frac{9}{3}$ ) to both sides gives $y = -\frac{2}{3}x - \frac{2}{3} + \frac{9}{3}$ , which is $y = -\frac{2}{3}x + \frac{7}{3}$ .
What is the equation of the line that passes through the point $(4, -2)$ and is perpendicular to the line $y = \frac{1}{3}x + 5$ ?
Answer: $y = -3x + 10$ The slope of the given line $y = \frac{1}{3}x + 5$ is $\frac{1}{3}$ . The slope of a line perpendicular to it is the negative reciprocal, which is $-3$ . Using the point-slope form $y - y_1 = m(x - x_1)$ with the point $(4, -2)$ and slope $-3$ , we get $y - (-2) = -3(x - 4)$ . This simplifies to $y + 2 = -3x + 12$ . Subtracting 2 from both sides gives the final equation $y = -3x + 10$ .
Based on the passage, which equation models the water level, $h$ , in centimeters as a function of time, $t$ , in minutes?
Answer: $h = \frac{3}{4}t + \frac{7}{2}$ The problem provides two points: $(2, 5)$ and $(6, 8)$ . First, calculate the slope (rate): $m = \frac{8 - 5}{6 - 2} = \frac{3}{4}$ . Next, use the point-slope form with the point $(2, 5)$ : $h - 5 = \frac{3}{4}(t - 2)$ . Distribute the slope: $h - 5 = \frac{3}{4}t - \frac{6}{4}$ , which simplifies to $h - 5 = \frac{3}{4}t - \frac{3}{2}$ . Finally, add 5 to both sides: $h = \frac{3}{4}t - \frac{3}{2} + 5$ . To combine the constants, use a common denominator: $h = \frac{3}{4}t - \frac{3}{2} + \frac{10}{2}$ , which simplifies to $h = \frac{3}{4}t + \frac{7}{2}$ .
A line is perpendicular to the line $y = -5$ and passes through the point $(3, 8)$ . What is the equation of this line?
Answer: $x = 3$ The line $y = -5$ is a horizontal line with a slope of 0. A line perpendicular to a horizontal line is a vertical line. A vertical line has an undefined slope and its equation is of the form $x = k$ , where $k$ is the x-coordinate of every point on the line. Since the line must pass through the point $(3, 8)$ , its equation must be $x = 3$ .
What is the equation of a line that is perpendicular to $x + 4y = 8$ and has the same x-intercept as this line?
Answer: $y = 4x - 32$ First, find the slope of the given line. $x + 4y = 8 \implies 4y = -x + 8 \implies y = -\frac{1}{4}x + 2$ . The slope is $-\frac{1}{4}$ . The slope of a perpendicular line is the negative reciprocal, which is $4$ . Next, find the x-intercept of the original line by setting $y=0$ : $x + 4(0) = 8 \implies x=8$ . So the new line must pass through the point $(8, 0)$ . Using the point-slope form with this point and the perpendicular slope: $y - 0 = 4(x - 8)$ , which simplifies to $y = 4x - 32$ .
A line passes through the points $(-3, 7)$ and $(5, 7)$ . What is the equation of this line?
Answer: $y = 7$ Since the y-coordinates of both points are the same (both are 7), the line is a horizontal line. The equation of a horizontal line is $y = k$ , where $k$ is the constant y-value. Therefore, the equation of this line is $y = 7$ . Alternatively, one could calculate the slope: $m = \frac{7 - 7}{5 - (-3)} = \frac{0}{8} = 0$ . Using point-slope form: $y - 7 = 0(x - 5)$ , which simplifies to $y - 7 = 0$ , or $y = 7$ .
Based on the passage, which equation models the profit, $P$ , in terms of the number of years, $t$ , after 2018?
Answer: $P = 5000t + 15000$ Let $t=0$ represent the year 2018. This gives us the point $(0, 15000)$ , meaning the y-intercept is 15,000. The year 2022 corresponds to $t=4$ , which gives the point $(4, 35000)$ . The slope (rate of change) is $m = \frac{35000 - 15000}{4 - 0} = \frac{20000}{4} = 5000$ . Using the slope-intercept form $P = mt + b$ , we get $P = 5000t + 15000$ .
Based on the passage, which equation relates her depth, $d$ , in feet to the time, $t$ , in minutes since she started ascending?
Answer: $d = -15t + 120$ The starting depth gives the y-intercept (the depth at $t=0$ ). So, the point is $(0, 120)$ , and $b=120$ . After 3 minutes, the depth is 75 feet, which gives the point $(3, 75)$ . The slope (rate of change in depth) is $m = \frac{75 - 120}{3 - 0} = \frac{-45}{3} = -15$ feet per minute. The negative sign indicates the depth is decreasing. Using the slope-intercept form $d = mt + b$ , the equation is $d = -15t + 120$ .
What is the equation of the line that is parallel to the x-axis and passes through the point $(-2, 6)$ ?
Answer: $y = 6$ The x-axis is a horizontal line with the equation $y = 0$ . Any line parallel to the x-axis must also be a horizontal line, and its equation will be of the form $y = k$ , where $k$ is the y-coordinate for every point on the line. Since the line passes through $(-2, 6)$ , the y-coordinate for all points must be 6. Therefore, the equation is $y = 6$ .
A line is parallel to $y = 3x - 5$ and has the same y-intercept as the line $2y + 4x = 12$ . What is the equation of this line?
Answer: $y = 3x + 6$ First, determine the required slope. The line is parallel to $y = 3x - 5$ , so it must have the same slope, $m = 3$ . Next, determine the required y-intercept. It is the same as the y-intercept of $2y + 4x = 12$ . Convert this equation to slope-intercept form: $2y = -4x + 12 \implies y = -2x + 6$ . The y-intercept is $b = 6$ . Finally, combine the slope $m=3$ and y-intercept $b=6$ to form the new equation: $y = 3x + 6$ .
A line passes through points $(1, 1)$ and $(4, -5)$ . Which of the following equations represents this line in standard form, $Ax + By = C$ ?
Answer: $2x + y = 3$ First, find the slope of the line: $m = \frac{-5 - 1}{4 - 1} = \frac{-6}{3} = -2$ . Using the point-slope form with point $(1, 1)$ , we have $y - 1 = -2(x - 1)$ . Distributing gives $y - 1 = -2x + 2$ . To convert to standard form $Ax + By = C$ , move the x-term to the left side and the constant term to the right side: $2x + y = 2 + 1$ , which simplifies to $2x + y = 3$ .

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