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How does using the point - slope form of a linear equation make it easier to write the equation of a line?
If given simply the slope of a line and a point through which it passes, and then told to find the equation of the line, one of the easiest ways of doing so is to use the point-slope formula.
How does using the point slope form a linear equation make it easier to write the equation of a line?
If given simply the slope of a line and a point through which it passes, and then told to find the equation of the line, one of the easiest ways of doing so is to use the point-slope formula.
How does using the point slope form of a linear equation make it easier to write the equation of a line?
If given simply the slope of a line and a point through which it passes, and then told to find the equation of the line, one of the easiest ways of doing so is to use the point-slope formula.
Find the equation of the line using the point-slope formula. Write the final equation using the slope-intercept form. perpendicular to 3y x and minus 4 and passes through the point (-21)?
To find the equation of a line that is perpendicular to the line given by (3y = x - 4), we first need to determine the slope of that line. Rearranging it into slope-intercept form (y = mx + b), we find the slope (m = \frac{1}{3}). The slope of the perpendicular line will be the negative reciprocal, which is (-3). Using the point-slope formula (y - y_1 = m(x - x_1)) with the point ((-2, 1)) and slope (-3), the equation becomes (y - 1 = -3(x + 2)). Simplifying this gives us (y = -3x - 5) in slope-intercept form.
How the slope of a curved line at a point can be found?
The slope of a curved line at a point is the slope of the tangent to the curve at that point. If you know the equation of the curve and the curve is well behaved, you can find the derivative of the equation of the curve. The value of the derivative, at the point in question, is the slope of the curved line at that point.
How does using the point slope form a linear equation make it easier to write the equation of a line?
If given simply the slope of a line and a point through which it passes, and then told to find the equation of the line, one of the easiest ways of doing so is to use the point-slope formula.
How does using the point - slope form of a linear equation make it easier to write the equation of a line?
If given simply the slope of a line and a point through which it passes, and then told to find the equation of the line, one of the easiest ways of doing so is to use the point-slope formula.
How does using the point-slope form of a linear equation make it easier to write the equation of a line?
If given simply the slope of a line and a point through which it passes, and then told to find the equation of the line, one of the easiest ways of doing so is to use the point-slope formula.
How does using the point slope form of a linear equation make it easier to write the equation of a line?
If given simply the slope of a line and a point through which it passes, and then told to find the equation of the line, one of the easiest ways of doing so is to use the point-slope formula.
How do you find an equation with a given slope?
Use point-slope formula
Find the equation of the line using the point-slope formula. Write the final equation using the slope-intercept form. perpendicular to 3y x and minus 4 and passes through the point (-21)?
To find the equation of a line that is perpendicular to the line given by (3y = x - 4), we first need to determine the slope of that line. Rearranging it into slope-intercept form (y = mx + b), we find the slope (m = \frac{1}{3}). The slope of the perpendicular line will be the negative reciprocal, which is (-3). Using the point-slope formula (y - y_1 = m(x - x_1)) with the point ((-2, 1)) and slope (-3), the equation becomes (y - 1 = -3(x + 2)). Simplifying this gives us (y = -3x - 5) in slope-intercept form.
How the slope of a curved line at a point can be found?
The slope of a curved line at a point is the slope of the tangent to the curve at that point. If you know the equation of the curve and the curve is well behaved, you can find the derivative of the equation of the curve. The value of the derivative, at the point in question, is the slope of the curved line at that point.
How do you find point slope form of a line?
By using the equation of a straight line y = mx+b whereas m is the slope of the line and b is the y intercept
Find equation perpendicular to given line contain given point?
If you know the slope of the line that your equation is perpendicular too, you find the negative reciprocal of it and use it as the slope for the line. (negative reciprocal = flip the slope over and change its sign. Ex: a slope of 2 has a negative reciprocal of -1/2. ) Then you use the given point, and put your equation in point-slope form. The general equation for point slope form is Y-y1=m(x-x1) The y1 is the y coordinate of the given point. X1 is the x coordinate of the given point. M is the slope that you found earlier. You now have your equation. If you are asked to put it in slope intercept form, simply distribute the numbers and solve the equation for y.
Find an equation of the line containing the given point and having the given slope 3-6m1?
Assuming the point is (3, -6) and the slope 1, the equation is x - y - 9 = 0
What is an equation of the line that passes through the point 50 and is perpendicular to the line 5x 6y24 express your answer in slope-intercept form?
Here is how to solve it. First, find the slope of the given line. To do this, solve the equation for "y". That will convert the equation to the slope-intercept form. From there, you can immediately read off the slope. Since parallel lines have the same slope, the line you are looking for will have the same slope. Now you need to use the point-slope form of the equation, with the given point, and the slope you just calculated. Finally, solve this equation for "y" to bring it into the requested slope-intercept form.
How is the normal drawn?
The normal line at a point on a surface is drawn perpendicular to the tangent line at that point. To find it, you first determine the slope of the tangent line by calculating the derivative of the function at that point. The slope of the normal line is the negative reciprocal of the tangent line's slope. Finally, you use the point-slope form of a linear equation to draw the normal line using the calculated slope and the coordinates of the point.
