Example Question - equation of a line
Here are examples of questions we've helped users solve.
Finding the Equation of a Line with a Given Point and Angle
<p>The equation of a line in point-slope form is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the line and \((x_1, y_1)\) is a point on the line.</p> <p>We are given the point \((1, 2)\) and that the line makes a \(30^\circ\) angle with the x-axis. The slope \(m\) of the line is the tangent of the angle it makes with the x-axis, thus \(m = \tan(30^\circ) = \frac{\sqrt{3}}{3}\).</p> <p>Substituting the point and the slope into the point-slope form equation, we get:</p> <p>\(y - 2 = \frac{\sqrt{3}}{3}(x - 1)\)</p> <p>Multiplying both sides by 3 to clear the fraction, we get:</p> <p>\(3(y - 2) = \sqrt{3}(x - 1)\)</p> <p>Expanding, we have:</p> <p>\(3y - 6 = \sqrt{3}x - \sqrt{3}\)</p> <p>Adding 6 and \(\sqrt{3}\) to both sides gives us the final equation:</p> <p>\(3y = \sqrt{3}x + 6 - \sqrt{3}\)</p> <p>Or we can express it in standard form \(Ax + By = C\) by rearranging terms:</p> <p>\(\sqrt{3}x - 3y = -6 + \sqrt{3}\)</p>
Finding the Equation of a Line Given Slope and Point
<p>The point-slope form of a line's equation is \( y - y_1 = m(x - x_1) \) where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line.</p> <p>For the given slope \( m = 2 \) and point \( (1, 3) \), we have:</p> <p>\( y - 3 = 2(x - 1) \)</p> <p>\( y - 3 = 2x - 2 \)</p> <p>\( y = 2x + 1 \)</p>
Finding the Equation of a Line Given Slope and a Point
Given slope, \( m = 2 \), and a point, \( (1,3) \), use the point-slope form of the equation of a line: \( y - y_1 = m(x - x_1) \). Substitute \( m = 2 \), \( x_1 = 1 \), and \( y_1 = 3 \) into the equation: \( y - 3 = 2(x - 1) \) Now, simplify and put it in slope-intercept form, \( y = mx + b \): \( y - 3 = 2x - 2 \) \( y = 2x + 1 \)
Finding the Equation of a Line with a Given Slope and Point
\[ \text{Given slope (m)} = 2, \text{ and point } (x_1, y_1) = (1, 3) \] \[ \text{Use point-slope form: } y - y_1 = m(x - x_1) \] \[ y - 3 = 2(x - 1) \] \[ y - 3 = 2x - 2 \] \[ y = 2x + 1 \]
Finding the Equation of a Line Given Slope and a Point
Given, slope \( m = \frac{2}{3} \) and point \( (1, -2) \). The equation of a line in slope-intercept form is \( y = mx + b \). Substituting the given slope and point into the equation to find \( b \): \( -2 = \left( \frac{2}{3} \right)(1) + b \) To solve for \( b \): \( b = -2 - \frac{2}{3} \) \( b = -\frac{6}{3} - \frac{2}{3} \) \( b = -\frac{8}{3} \) The equation of the line is: \( y = \frac{2}{3}x - \frac{8}{3} \)
Graphing a Line from Point-Slope Form Equation
To graph the line given by the equation \( y - 2 = \frac{-2}{5}(x - 4) \), you can follow these steps: 1. Recognize the equation form: The equation is in point-slope form, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope of the line. 2. Identify the slope and a point: From the given equation, the slope (\( m \)) is \( -\frac{2}{5} \) and the point (\( x_1, y_1 \)) is (4, 2). 3. Plot the given point: Start by plotting the point (4, 2) on the coordinate plane. This is a point through which the line will pass. 4. Use the slope to find another point: The slope \( -\frac{2}{5} \) means that for every 5 units you move to the right (in the positive x-direction), you move 2 units down (in the negative y-direction) because the slope is negative. You could also move 5 units to the left (negative x-direction) and 2 units up (positive y-direction) to find another point. 5. Plot the second point: For example, starting at point (4, 2) and moving 5 units to the right, you reach the point (9, 0). Plot this point on the coordinate plane. 6. Draw the line: Once you have these two points plotted, you can use a ruler to draw a straight line through them, extending it across the coordinate plane. This line represents all the points that satisfy the equation. Here's a rough step-by-step description of plotting the points and drawing the line: - Place a dot at (4, 2). - From (4, 2), move 5 units to the right to get to (9, 2). - From (9, 2), move 2 units down to get to (9, 0). - Place a dot at (9, 0). - Connect the two dots with a straight line. - Extend the line through and beyond both points to indicate that it continues infinitely in both directions. If you're graphing this by hand or using a graphing tool, make sure to label your axes and scale appropriately to accurately depict the slope and points.
Graphing a Line in Slope-Intercept Form
Based on the image provided, the equation of the line is \(y + 3 = \frac{9}{4}(x + 4)\). To graph this line, it's typically easiest to write it in slope-intercept form, which is \(y = mx + b\), where \(m\) is the slope of the line and \(b\) is the y-intercept. Let's rewrite the given equation in slope-intercept form: \[ \begin{align*} y + 3 &= \frac{9}{4}(x + 4) \\ y &= \frac{9}{4}x + \frac{9}{4} \cdot 4 - 3 \\ y &= \frac{9}{4}x + 9 - 3 \\ y &= \frac{9}{4}x + 6 \end{align*} \] Now we have the equation in slope-intercept form, with a slope \(m=\frac{9}{4}\) and a y-intercept \(b=6\). To graph this line: 1. Start at the y-intercept (0,6) on the graph. 2. Use the slope to find another point. The slope is the rise over the run, meaning you go up 9 units and right 4 units from the y-intercept (or down 9 and left 4, for a point in the opposite direction). 3. Plot the second point. 4. Draw a line through the two points, extending it across the graph. This straight line represents the equation \(y = \frac{9}{4}x + 6\).
Equation of a Line Passing Through Two Points
Câu trả lời của bạn cho câu hỏi từ hình ảnh được cung cấp là như sau: Câu 22: Chúng ta cần tìm phương trình của đường thẳng qua hai điểm A(1;2;-3) và B(3;1;-1). Ta biết rằng vectơ chỉ phương của đường thẳng đi qua hai điểm A và B có thể tính được bằng cách lấy tọa độ của điểm B trừ cho tọa độ của điểm A. Ta có: \(\overrightarrow{AB} = (3 - 1, 1 - 2, -1 - (-3)) = (2, -1, 2)\). Vectơ chỉ phương của đường thẳng \( \overrightarrow{u} = (2, -1, 2) \). Phương trình tham số của đường thẳng là: \(x = x_0 + at\), \(y = y_0 + bt\), và \(z = z_0 + ct\), trong đó \((x_0, y_0, z_0)\) là một điểm trên đường thẳng, và \((a, b, c)\) là vectơ chỉ phương của đường thẳng. Do đó, phương trình tham số của đường thẳng đi qua hai điểm A và B là: \(x = 1 + 2t\), \(y = 2 - t\), và \(z = -3 + 2t\). Chuyển các phương trình về dạng phương trình đường thẳng tổng quát tham số t, ta được: \(\frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z + 3}{2}\) Đây chính là phương án A.
Analyzing a Line on a Coordinate Plane
La imagen muestra un plano de coordenadas con una línea recta dibujada a través de él. Para resolver la pregunta que se haga referencia a esta línea recta, necesito que me digas específicamente qué necesitas resolver. ¿Estás buscando encontrar la ecuación de la línea recta, identificar puntos particulares a lo largo de la línea, o algún otro tipo de análisis? Por favor proporciona más detalles.
Finding Equation of a Line and Inequalities
To find the equation of the line, you need to find its slope and y-intercept. From the graph, you can see the line crosses the y-axis at the point (0, 1), which means the y-intercept is 1. To find the slope, you can pick two points on the line. The line passes through the points (0, 1) and (-2, -1) which can be clearly seen on the graph. The slope (m) is calculated as the rise over the run: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 1}{-2 - 0} = \frac{-2}{-2} = 1. \] So, the slope is 1. With a slope of 1 and a y-intercept of 1, the equation of the line in slope-intercept form (y = mx + b) is: \[ y = 1x + 1 \] or simply: \[ y = x + 1. \] For inequalities: a) Since the line is solid, this means that the inequality includes equality; that is, points on the line satisfy the inequality. Since the area below the line is shaded, the inequality represents all points where y is less than or equal to x + 1. Therefore, the inequality for part a) is: \[ y \leq x + 1. \] b) and c) are not provided explicitly in the question, but typically they would be related to different shading regions if provided. If the shaded region was above the line for part b), for example, the inequality would be y ≥ x + 1. If the shaded region for part c) omitted points on the line, the inequality would use strict inequality signs (< or >) rather than ≤ or ≥.
Equation of a Line and Inequalities in Shaded Regions
I'll help you to determine the equation of the line shown in part a of the image, and then we can use this information to write inequalities for the shaded regions shown in parts b and c. To find the equation of a line, we need to identify two key components: the slope and the y-intercept. In the coordinate grid provided, the line crosses the y-axis at (0, -2), which means the y-intercept is -2. The line also looks as if it passes through another point with integer coordinates, which we can use to determine the slope. I see that the line goes through (2, 2) along with (0, -2). The slope (\(m\)) is the change in y over the change in x when moving from one point to another: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-2)}{2 - 0} \] \[ m = \frac{4}{2} = 2 \] So the slope of the line is 2. Now that we have both the slope and the y-intercept, we can write the equation of the line in slope-intercept form (y = mx + b): \[ y = 2x - 2 \] For part b, if the graph is shaded above this line, then the inequality representing the shaded region would use a "greater than" symbol, because the y-values are greater than the y-values on the line for any given x-value. The inequality for the graph in part b would be: \[ y > 2x - 2 \] For part c, if the graph is shaded below this line, then the inequality representing the shaded region would use a "less than" symbol, because the y-values are less than the y-values on the line for any given x-value. The inequality for the graph in part c would be: \[ y < 2x - 2 \]
Determining Equations of Lines and Inequalities
To find the equation of the line in part a, we need to determine the slope and the y-intercept. The equation of a line in slope-intercept form is: y = mx + b where m is the slope of the line, and b is the y-intercept. We can find the slope by looking at two points on the line and using the slope formula: slope (m) = (change in y) / (change in x) From the image, we can pick two points that the line passes through. Let's choose (0, -2) and (2, -1) since they're clearly on the grid intersections. Now, we find the slope: m = (y2 - y1) / (x2 - x1) m = (-1 - (-2)) / (2 - 0) m = (1) / (2) m = 1/2 Next, we find the y-intercept (b). This is where the line crosses the y-axis. Looking at the graph, we can see that this occurs at (0, -2), so b = -2. Now we have the slope m = 1/2 and y-intercept b = -2, the equation of the line is: y = (1/2)x - 2 To find the inequalities for parts b and c, we would normally look for instructions that designate whether the area above or below the line should be shaded. Since the image only shows the line and does not specify the inequalities or shaded regions, we can't determine what the inequalities would be. However, if you are being asked for the inequality that includes points below the line, the inequality would be: y ≤ (1/2)x - 2 If the inequality should include points above the line, it would be: y ≥ (1/2)x - 2 Remember, without additional information or context regarding which side of the line should be considered for the inequality, you cannot definitively state the inequality.
Finding Parallel Lines by Comparing Slopes
To find a line that is parallel to the given line, you need to ensure that it has the same slope. The given line is \(2y = 3x - 1\). We can rearrange this into slope-intercept form \(y = mx + b\) to find the slope \(m\). Divide both sides of the equation by 2 to get the slope-intercept form: \[ y = \frac{3}{2}x - \frac{1}{2} \] Here, \(m = \frac{3}{2}\) is the slope. Now we need to find the equation from the options that has the same slope of \(\frac{3}{2}\). Let's check each option by rewriting them in the slope-intercept form if necessary: a. \(4y = 6x + 8\): Divide by 4 to get \(y = \frac{6}{4}x + 2\), which simplifies to \(y = \frac{3}{2}x + 2\). The slope here is \(\frac{3}{2}\), which is the same as the given line's slope. This line is thus parallel to the given line. b. \(3y = 2x - 3\): Divide by 3 to get \(y = \frac{2}{3}x - 1\). The slope here is \(\frac{2}{3}\), which is not equal to \(\frac{3}{2}\). c. \(2y = x - 3\): Divide by 2 to get \(y = \frac{1}{2}x - \frac{3}{2}\). The slope here is \(\frac{1}{2}\), which is not equal to \(\frac{3}{2}\). d. \(y = \frac{1}{3}x - 1\): The slope here is \(\frac{1}{3}\), which is not equal to \(\frac{3}{2}\). The correct answer is option a, as it's the only equation with a slope of \(\frac{3}{2}\), which means it is parallel to the given line.
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