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If you mean a slope of -2 passing through (5, 0) then the equation is y = -2x+10

Q: What equation has a slope of -2 and passes through (50)?

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Two lines are parallel if they have the same slope. First we need to find the slope m of the line with equation y - x = 6. So, y - x = 6 add x to both sides to turn the equation into the slope-intercept form y = x + 6 We know that the slope is the coefficient of the variable x, so the slope is 1. So the line that passes through (5,0) and parallel to the given line has also slope 1. Using the point (5, 0) = (x1, y1) and m = 1, we write the point-slope form of the equation of the required line, (y - y1) = m(x - x1) substitute the known values (y - 0) = 1(x - 5) y = x + 5 subtract x to both sides to turn this equation into the general form y - x = 5 Thus, the line that passes through (5,0) and parallel to the given line has the equation y - x = 5.

Without an equality sign and not knowing some of the plus or minus values of the given terms it can't be considered to be a straight line equation. But if you mean: y = 12x-3 a given point of (5, 0) Then the perpendicular equation is: y = -1/12x+5/12 whereas -1/12 is the slope and 5/12 is the y intercept

Gradient of given line is 2.1/5 = 21/50 So gradient (slope) of perpendicular is -50/21 = -2.381 (to 3 dp)

Yes. This represents a line with a slope of 1 which intersects the axis at {l:50, w:0} and {l:0, w:50}

If you mean points of: (5, 0) and (6, 2) then the slope works out as 2

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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.

If you mean points of (-5, 0) and (10, 9) then the slope is 3/5 and the straight line equation is 5y = 3x+15

5x - 3y - 50 is an expression NOT an equation. It is not possible to rewrtite an expression in a slope intercept form.However, IF the equation were 5x - 3y - 50 = 0then 3y = 5x - 50so thaty = 5/3x - 50/3 is the equation in the slope - intercept from.

Two lines are parallel if they have the same slope. First we need to find the slope m of the line with equation y - x = 6. So, y - x = 6 add x to both sides to turn the equation into the slope-intercept form y = x + 6 We know that the slope is the coefficient of the variable x, so the slope is 1. So the line that passes through (5,0) and parallel to the given line has also slope 1. Using the point (5, 0) = (x1, y1) and m = 1, we write the point-slope form of the equation of the required line, (y - y1) = m(x - x1) substitute the known values (y - 0) = 1(x - 5) y = x + 5 subtract x to both sides to turn this equation into the general form y - x = 5 Thus, the line that passes through (5,0) and parallel to the given line has the equation y - x = 5.

Without an equality sign and not knowing some of the plus or minus values of the given terms it can't be considered to be a straight line equation. But if you mean: y = 12x-3 a given point of (5, 0) Then the perpendicular equation is: y = -1/12x+5/12 whereas -1/12 is the slope and 5/12 is the y intercept

Without an equality sign and not knowing the plus or minus value of 5y it is not an equation.

Gradient of given line is 2.1/5 = 21/50 So gradient (slope) of perpendicular is -50/21 = -2.381 (to 3 dp)

It means that the current passes through 0 50 times a second.

Yes. This represents a line with a slope of 1 which intersects the axis at {l:50, w:0} and {l:0, w:50}

If you mean points of: (5, 0) and (6, 2) then the slope works out as 2

50%

The graph is a straight line, with a slope of 40, passing through the point y=50 on the y-axis.