Q: What is the equation of a line that passes through points -1 2 and 5 2?

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The equation for the given points is y = x+4 in slope intercept form

If you mean points of (2, -2) and (-4, 22) then the equation is y = -4x+6

Answer this question…y = 2x + 6

y = 3x-4

Points: (-1, 2) and (5, 2) Slope: 0 Equation: y = 2

Related questions

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

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

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

3

Answer this question…y = 2x + 6

Slope-intercept form

Points: (0, -2) and (6, 0) Slope: 1/3 Equation of line: 3y = x-6