Write the equation of the line that passes through the points (3, -5) and (-4, -5)
Y= -3x + 8
Points: (-1, 7) and (-2, 3) Slope: 4 Equation: y = 4x+11
Points: (2, 5) and (4, 3) Slope: -1 Equation: y = -x+7
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
Write the equation of the line that passes through the points (3, -5) and (-4, -5)
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
Y= -3x + 8
It is y = 2.
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.
Points: (0. 5) and (2, 3) Slope: -1 Equation: y = -x+5
If you mean points of (-4, 2) and (4, -2) Then the straight line equation works out as 2y = -x
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.
Answer this question…y = 2x + 6
If the points are (1,5) and (0,0) y = 5x