Points: (-4, 1) and (4, -6)
Slope: -7/8
Equation: y = -7/8x-5/2 or as 7x+8y+20 = 0
A line that is parallel to the y-axis is a vertical line. The equation of a vertical line is of the form ( x = k ), where ( k ) is a constant. Since the line passes through the points ( (4, y) ) and ( (3, y) ), the line that is parallel to the y-axis and passes through these points would have the equation ( x = 4 ) or ( x = 3 ), depending on which point you choose.
It is y = 2.
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
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
It is y = 2.
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
3
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
Slope-intercept form