Example Question - point-slope form
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>
Equation of a Parallel Line Passing Through a Given Point
Como la pregunta es sobre matemáticas, procedo como Profesor de Matemáticas para abordar el problema. La pregunta requiere que se encuentre la ecuación de una recta que pase por un punto dado y que sea paralela a una recta dada. La recta dada es \( \frac{2y-3}{x-1} = 3 \). Para encontrar la pendiente, primero reescribimos la ecuación en la forma \( y = mx + b \), donde \( m \) es la pendiente. \[ \frac{2y-3}{x-1} = 3 \rightarrow 2y - 3 = 3(x - 1) \rightarrow 2y = 3x - 3 + 3 \] \[ 2y = 3x \rightarrow y = \frac{3}{2}x \] Por lo tanto, la pendiente de la recta dada es \( \frac{3}{2} \). Como la recta que buscamos debe ser paralela, tendrá la misma pendiente \( \frac{3}{2} \). La ecuación de la recta con pendiente \( \frac{3}{2} \) que pasa por el punto \( (-2, -1) \) se encontrará utilizando la forma punto-pendiente de la ecuación de una recta: \( y - y_1 = m(x - x_1) \), donde \( (x_1, y_1) \) es un punto en la recta y \( m \) es la pendiente. \[ y - (-1) = \frac{3}{2}(x - (-2)) \rightarrow y + 1 = \frac{3}{2}(x + 2) \] \[ \Rightarrow y + 1 = \frac{3}{2}x + 3 \] Restando 1 a ambos lados nos da la ecuación final de la recta: \[ y = \frac{3}{2}x + 2 \] Esta es la ecuación de la recta que pasa por el punto \( (-2, -1) \) y es paralela a la recta dada \( \frac{2y-3}{x-1} = 3 \).
Parallel Line Equations
La primera pregunta es encontrar la ecuación de una recta que pasa por el punto \( (\frac{1}{2}, \frac{1}{3}) \) y tiene pendiente \( m = -2 \). <p>\( y - y_1 = m(x - x_1) \)</p> <p>\( y - \frac{1}{3} = -2(x - \frac{1}{2}) \)</p> <p>\( y - \frac{1}{3} = -2x + 1 \)</p> <p>\( y = -2x + 1 + \frac{1}{3} \)</p> <p>\( y = -2x + \frac{4}{3} \)</p> La segunda pregunta es encontrar la ecuación de una recta que pasa por el punto \( (1, 0) \) y es paralela a la recta \( y = 2x - 3 \). <p>Dado que la recta es paralela, tiene la misma pendiente, \( m = 2 \).</p> <p>\( y - y_1 = m(x - x_1) \)</p> <p>\( y - 0 = 2(x - 1) \)</p> <p>\( y = 2x - 2 \)</p> La tercera pregunta es encontrar la ecuación de una recta que pasa por el punto \( (-2, -1) \) y es paralela a la recta \( \frac{y-1}{2} = \frac{2x-3}{3} \). <p>Primero, encontramos la pendiente de la recta dada resolviendo para \( y \).</p> <p>\( \frac{y-1}{2} = \frac{2x-3}{3} \)</p> <p>\( 3(y - 1) = 2(2x - 3) \)</p> <p>\( 3y - 3 = 4x - 6 \)</p> <p>\( 3y = 4x - 3 \)</p> <p>\( y = \frac{4}{3}x - 1 \)</p> <p>Entonces, la pendiente de la recta a encontrar es \( m = \frac{4}{3} \).</p> <p>Usamos la forma punto-pendiente para encontrar la ecuación de la nueva recta.</p> <p>\( y - y_1 = m(x - x_1) \)</p> <p>\( y + 1 = \frac{4}{3}(x + 2) \)</p> <p>\( y + 1 = \frac{4}{3}x + \frac{8}{3} \)</p> <p>\( y = \frac{4}{3}x + \frac{8}{3} - 1 \)</p> <p>\( y = \frac{4}{3}x + \frac{5}{3} \)</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
\[ \text{Given slope (m)} = 2, \text{ and point } (x_1, y_1) = (1, 3) \] \[ \text{Use point-slope form: } y - y_1 = m(x - x_1) \] \[ y - 3 = 2(x - 1) \] \[ y - 3 = 2x - 2 \] \[ y = 2x + 1 \]
Equation of Perpendicular Line
The equation of line p is given as y = -4x + 1. This equation is in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. For line p, the slope is -4. Line q is supposed to be perpendicular to line p. When two lines are perpendicular, the product of their slopes is -1. This means that if the slope of line p is m, then the slope of line q will be -1/m. Therefore, the slope of line q is -1/(-4) = 1/4. Now that we know the slope of line q is 1/4, we can use the point it passes through, (-6, 1), to find the y-intercept (b) of line q. Starting with the point-slope form of the line equation: y - y1 = m(x - x1) Plugging in the slope (m = 1/4) and the point (-6, 1): y - 1 = 1/4(x - (-6)) y - 1 = 1/4(x + 6) Now, distribute 1/4 to (x + 6): y - 1 = 1/4x + 1/4(6) Simplify: y - 1 = 1/4x + 6/4 y - 1 = 1/4x + 3/2 Finally, we want to write this in slope-intercept form, so we solve for y by adding 1 to both sides: y = 1/4x + 3/2 + 1 Since we want to write numbers as simplified fractions or integers, let's convert 1 to a fraction with a denominator of 2: y = 1/4x + 3/2 + 2/2 Combine the fractions: y = 1/4x + 5/2 And so, the equation of line q in slope-intercept form is: y = 1/4x + 5/2
Finding Equation of Perpendicular Line through Point
To find the equation of the line h, which is perpendicular to line g and passes through the point (3, -9), we need to start by identifying the slope of line g. The equation for line g is written in slope-intercept form as: \[ y = \frac{3}{10}x - 8 \] The slope (m) of line g is the coefficient of x, which is \(\frac{3}{10}\). For a line to be perpendicular to another line, its slope must be the negative reciprocal of the original line's slope. The negative reciprocal of \(\frac{3}{10}\) is \(-\frac{10}{3}\) (because when you flip the fraction and change the sign, you get the negative reciprocal). Now that we have the slope of the perpendicular line, we can use the point-slope form of the equation of a line to find line h's equation. Point-slope form is: \[ y - y_1 = m(x - x_1) \] where \(m\) is the slope and \((x_1, y_1)\) is a point on the line. Using the slope of line h, which is \(-\frac{10}{3}\), and the given point (3, -9), we plug these into the point-slope form: \[ y - (-9) = -\frac{10}{3}(x - 3) \] Simplify the equation: \[ y + 9 = -\frac{10}{3}x + \frac{10}{3} \cdot 3 \] \[ y + 9 = -\frac{10}{3}x + 10 \] We want to get it into slope-intercept form (y = mx + b), so we'll isolate y: \[ y = -\frac{10}{3}x + 10 - 9 \] \[ y = -\frac{10}{3}x + 1 \] The equation of line h in slope-intercept form, with numbers as simplified proper fractions, improper fractions, or integers, is: \[ y = -\frac{10}{3}x + 1 \]
Equation of Perpendicular Line Passing Through a Point
To find the equation of the line h, which is perpendicular to line g, and passes through the point (3, −9), we first need to determine the slope of line g. The equation for line g is given by: \[ y = \frac{3}{10}x - \frac{8}{10} \] The slope of line g is the coefficient of x, which is \(\frac{3}{10}\). Two lines that are perpendicular to each other have slopes that are negative reciprocals of one another. So, the slope (m) of line h will be: \[ m = -\frac{1}{\text{slope of g}} = -\frac{1}{\frac{3}{10}} = -\frac{10}{3} \] Now we know the slope of line h, we can use the point-slope form of the equation of a line to find its equation. The point-slope form is: \[ y - y_1 = m(x - x_1) \] Plugging in our values for the slope and the coordinates of the given point (3, −9), we get: \[ y - (-9) = -\frac{10}{3}(x - 3) \] \[ y + 9 = -\frac{10}{3}x + 10 \] To write this in slope-intercept form, which is \( y = mx + b \), we solve for y: \[ y = -\frac{10}{3}x + 10 - 9 \] \[ y = -\frac{10}{3}x + 1 \] Here \( b \), the y-intercept, is 1. The numbers in the equation are already simplified proper fractions, improper fractions, or integers. So, the equation for line h in slope-intercept form is: \[ y = -\frac{10}{3}x + 1 \]
Finding Equation of Perpendicular Line
To find the equation of line h, which is perpendicular to line g, we will follow these steps: 1. Identify the slope of line g. 2. Determine the slope of line h. 3. Use the point-slope form to create the equation of line h. 4. Convert the equation into slope-intercept form. The equation for line g is given as \( y = \frac{3}{10}x - 8 \). The slope of line g is the coefficient of x, which is \( \frac{3}{10} \). Lines that are perpendicular to each other have slopes that are negative reciprocals. Therefore, if the slope of line g is \( \frac{3}{10} \), the slope of line h will be its negative reciprocal, which is \( -\frac{10}{3} \). Now, we know that line h has a slope of \( -\frac{10}{3} \) and it passes through the point (3, -9). We can use the point-slope form of the equation to find the equation of line h: \( y - y_1 = m(x - x_1) \) Substitute m (slope) with \( -\frac{10}{3} \) and \( (x_1, y_1) \) with (3, -9): \( y - (-9) = -\frac{10}{3}(x - 3) \) Simplify and solve for y to get the equation in slope-intercept form: \( y + 9 = -\frac{10}{3}x + 10 \) \( y = -\frac{10}{3}x + 10 - 9 \) \( y = -\frac{10}{3}x + 1 \) Therefore, the equation of line h in slope-intercept form is \( y = -\frac{10}{3}x + 1 \), with the numbers in the equation as simplified improper fractions or integers.
Finding a Parallel Line Passing Through a Point
The equation for line q is given as: \( y = -5 - \frac{1}{8}(x + 2) \) To find a line parallel to q that passes through the point (-6, 1), we need to keep the slope the same, since parallel lines have equal slopes. First, let's rewrite the equation for line q in slope-intercept form (\( y = mx + b \)), where \( m \) is the slope and \( b \) is the y-intercept. Rewrite the equation of line q to make the slope more apparent: \( y = -5 - \frac{1}{8}x - \frac{1}{8}(2) \) \( y = -5 - \frac{1}{8}x - \frac{1}{4} \) \( y = -\frac{1}{8}x - 5 - \frac{1}{4} \) \( y = -\frac{1}{8}x - 5.25 \) \( y = -\frac{1}{8}x - \frac{21}{4} \) Now we know the slope of line q is \( -\frac{1}{8} \). Since line r is parallel to line q, it will also have a slope of \( -\frac{1}{8} \). Using the point-slope form of a line's equation ( \( y - y_1 = m(x - x_1) \) ), where \( m \) is the slope and \( (x_1, y_1) \) is the point (-6, 1) through which line r passes, we can write: \( y - 1 = -\frac{1}{8}(x - (-6)) \) \( y - 1 = -\frac{1}{8}(x + 6) \) Now we can put this in slope-intercept form: \( y = -\frac{1}{8}x - \frac{1}{8}(6) + 1 \) \( y = -\frac{1}{8}x - \frac{3}{4} + 1 \) \( y = -\frac{1}{8}x + \frac{1}{4} \) So, the equation of line r in slope-intercept form is: \( y = -\frac{1}{8}x + \frac{1}{4} \)
Finding Equation of Parallel Line Passing Through a Point
To find the equation of line r that is parallel to line q and passes through the point (-6, 1), we start by determining the slope of line q. The equation of line q is given in slope-intercept form as: y = -5 - 1/2(x + 2) In slope-intercept form, which is y = mx + b, m represents the slope and b represents the y-intercept. Based on line q's equation, the slope (m) is -1/2. Since line r is parallel to line q, line r will have the same slope as line q. Therefore, the slope of line r will also be -1/2. Using the point-slope form of a line equation, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope, we can substitute the slope and the point through which line r passes. The point (-6, 1) will be our (x1, y1), and our slope (m) will be -1/2. y - y1 = m(x - x1) y - 1 = -1/2(x - (-6)) y - 1 = -1/2(x + 6) Now, we solve for y to get the equation in slope-intercept form. y = -1/2 * x - 1/2 * 6 + 1 y = -1/2 * x - 3 + 1 y = -1/2 * x - 2 So, the equation of line r in slope-intercept form is: y = -1/2x - 2 This equation uses numbers written as simplified proper fractions, improper fractions, or integers, as requested.
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 Using Point-Slope Form
To graph the line given by the equation \( y - 2 = \frac{-2}{5}(x - 4) \), you can follow these steps: 1. **Identify the slope and y-intercept:** The equation is in point-slope form, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line. In this case, the slope (\( m \)) is \( \frac{-2}{5} \) and the point on the line is (4, 2). 2. **Plot the point on the graph:** Begin by plotting the point (4, 2) on your graph paper. This will be a point through which the line passes. 3. **Use the slope to find another point:** The slope \( \frac{-2}{5} \) means you move 2 units down and 5 units to the right from the point (4, 2) to find another point on the line. You could also move 2 units up and 5 units to the left if you wish to go in the opposite direction. 4. **Draw the line:** Connect the two points with a straight line. This line will extend infinitely in both directions. It's important to be precise and to plot points correctly. If you have graph paper, use a ruler to ensure that the line passes through the points accurately and extends across the grid.
Graphing a Linear Equation in Point-Slope Form
The equation provided in the image is a linear equation in point-slope form, which is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope of the line. The given equation is: \[ y - 5 = -\frac{8}{3}(x - 6) \] From this equation, we can identify the slope \( m = -\frac{8}{3} \) and a point on the line \( (x_1, y_1) = (6, 5) \). To graph this line, follow these steps: 1. Plot the point \( (6, 5) \) on a coordinate plane. 2. Use the slope to find additional points. The slope \( -\frac{8}{3} \) means that for every 3 units you move to the right (positive direction along the x-axis), you move 8 units down (negative direction along the y-axis). 3. From \( (6, 5) \), move 3 units to the right to \( (9, 5) \) and then 8 units down to \( (9, -3) \). Plot this second point. 4. Draw a straight line through the two points to extend it across the graph. Now you have graphed the line based on the provided linear equation.
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