The points on the line y + .5x are all the points of the form (x, y) that satisfy the equation y + .5x = 0.
This is a standard form linear equation, where y is the dependent variable and x is the independent variable. The equation y + .5x = 0 can be rewritten as y = -.5x. This is the slope-intercept form of the equation of a line, where the slope of the line is -.5 and the y-intercept is 0.
Thus, the line y + .5x is a line with a slope of -.5 and a y-intercept of 0. All the points on this line have an x-coordinate that satisfies the equation y + .5x = 0.
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B.Sai_ Kiran
ChatGPT Dec 15
The equations will have the same slope as y = 5x+9 but a different y intercept
Slope: -5 Points: (6, 3) Equation: y = -5x+33
-5x + 6y = 30At the x-intercept, y=0:-5x = 30 ==> x = -6At the y-intercept, x=0:6y = 30 ==> y = 5
If you mean: y=-5x+10 and the point (3, 10) then the parallel equation is y=-5x+25
If you mean y = 2x+3 and y = -1/2x+4 then the two lines are perpendicular to each other meeting at right angles.
-4
If: y = -5x+3 then the points are (2, -7) and (6, -27)
-10
(6, 40) and (-1, 5)
It will be any of the equations that has the same slope of y = 5x+9 but with a different y intercept
The equations will have the same slope as y = 5x+9 but a different y intercept
Y = 5X + 1 Is a line and should pass the vertical line test for functions, so this is a function.
That is a linear equation in two variables, x and y. The solution consists of all points on the line y = -5x + 19.
-9
No, it's -5.
If: y = x^2 +4 and 5x+10-y = 0 or y = 5x+10 Then: x^2 +4 = 5x +10 or x^2-5x-6 = 0 Solving the above quadratic equation: x = -1 or x = 6 By substitution end points of the line are at: (6, 40) and (-1, 5) Length of line is the square root of: (-1-6)^2 plus (5-40)^2 = 36 rounded
y = 5x + 6At the x-intercept, y = 0 .5x + 6 = 05x = -6x = -6/5