Write the equation of the line that passes through the points (3, -5) and (-4, -5)
Y= -3x + 8
Points: (0, -2) and (6, 0) Slope: 1/3 Equation of line: 3y = x-6
Points: (6, -3) and (-4, -9) Slope: 3/5 Equation: 5y = 3x-33
Points: (-1, 7) and (-2, 3) Slope: 4 Equation: y = 4x+11
Write the equation of the line that passes through the points (3, -5) and (-4, -5)
Plug both points into the equation of a line, y =m*x + b and then solve the system of equations for m and b to get equation of the line through the points.
In order to find the equation of a tangent line you must take the derivative of the original equation and then find the points that it passes through.
If you mean points of (-4, 2) and (4, -2) Then the straight line equation works out as 2y = -x
Y= -3x + 8
It is y = 2.
If the line passes through (5, 2) and (5, 7), then the x value stays constant for those two points, and since it is a line, the x value stays constant for the whole line, so the equation of the line isx = 5
3
If you mean points of (2, -2) and (-4, 22) then the equation is y = -4x+6
The equation for the given points is y = x+4 in slope intercept form
Answer this question…y = 2x + 6
Slope-intercept form