4x + y + c = 0 or, for a line going through a given point (xo, yo): y + 4x - (xo + yo) = 0 The gradient of a line multiplied by the gradient of a line perpendicular to it is -1; or in other words: The gradient of the perpendicular line is the negative reciprocal of the gradient of the line. Thus: 2x - 8y + 23 = 0 ⇒ 8y = 2x + 23 ⇒ y = 1/4x + 23/8 ⇒ gradient of perpendicular line is -1 ÷ 1/4 = -4 Thus the equation of the perpendicular line to 2x - 8y + 23 = 0 is 4x + y + c = 0. To find the line through point (xo, yo) perpendicular to 2x - 8y + 23 = 0, use the format: y - yo = m(x - xo) ⇒ y - yo = -4(x - xo) ⇒ y + 4x - (xo + yo) = 0
When equation of line is y=-4x+3, Gradient is -4 (as seen from the coefficient of x) and the y-intercept is +3 (point where x=0)
Gradient = [-4 - 0]/[6 - (-2)] = 4/8 = 0.5 Angle of inclination = arctan(0.5) = 0.464 radians.
L1: y = 1/2*x - 3Gradient of the line = 1/2Negative reciprocal of gradient = -1/(1/2) = -2That is, gradient of perpendicular = -2.This line goes through (0,3),(y - 3) = 2*(x - 0)y - 3 = 2xy = 2x + 3L1: y = 1/2*x - 3Gradient of the line = 1/2Negative reciprocal of gradient = -1/(1/2) = -2That is, gradient of perpendicular = -2.This line goes through (0,3),(y - 3) = 2*(x - 0)y - 3 = 2xy = 2x + 3L1: y = 1/2*x - 3Gradient of the line = 1/2Negative reciprocal of gradient = -1/(1/2) = -2That is, gradient of perpendicular = -2.This line goes through (0,3),(y - 3) = 2*(x - 0)y - 3 = 2xy = 2x + 3L1: y = 1/2*x - 3Gradient of the line = 1/2Negative reciprocal of gradient = -1/(1/2) = -2That is, gradient of perpendicular = -2.This line goes through (0,3),(y - 3) = 2*(x - 0)y - 3 = 2xy = 2x + 3
The gradient of a line is the same as the slope of a line. It will tell someone measuring the line how straight the line is.
4x + y + c = 0 or, for a line going through a given point (xo, yo): y + 4x - (xo + yo) = 0 The gradient of a line multiplied by the gradient of a line perpendicular to it is -1; or in other words: The gradient of the perpendicular line is the negative reciprocal of the gradient of the line. Thus: 2x - 8y + 23 = 0 ⇒ 8y = 2x + 23 ⇒ y = 1/4x + 23/8 ⇒ gradient of perpendicular line is -1 ÷ 1/4 = -4 Thus the equation of the perpendicular line to 2x - 8y + 23 = 0 is 4x + y + c = 0. To find the line through point (xo, yo) perpendicular to 2x - 8y + 23 = 0, use the format: y - yo = m(x - xo) ⇒ y - yo = -4(x - xo) ⇒ y + 4x - (xo + yo) = 0
basically the reciprocal of the original lines gradient is going to be the gradient for the perpendicular line (remember the signs should switch). For example if i had a line with the gradient of 3, then the gradient of the perpendicular line will be -1over3. But if the line had the gradient of -3, then the line perpendicular to that line will have the gradient 1over3.
By definition, lines are parallel if they have the same gradient (slope). Any horizontal line has a gradient of 0, so it is parallel to any other horizontal line.
When equation of line is y=-4x+3, Gradient is -4 (as seen from the coefficient of x) and the y-intercept is +3 (point where x=0)
Gradient = [-4 - 0]/[6 - (-2)] = 4/8 = 0.5 Angle of inclination = arctan(0.5) = 0.464 radians.
L1: y = 1/2*x - 3Gradient of the line = 1/2Negative reciprocal of gradient = -1/(1/2) = -2That is, gradient of perpendicular = -2.This line goes through (0,3),(y - 3) = 2*(x - 0)y - 3 = 2xy = 2x + 3L1: y = 1/2*x - 3Gradient of the line = 1/2Negative reciprocal of gradient = -1/(1/2) = -2That is, gradient of perpendicular = -2.This line goes through (0,3),(y - 3) = 2*(x - 0)y - 3 = 2xy = 2x + 3L1: y = 1/2*x - 3Gradient of the line = 1/2Negative reciprocal of gradient = -1/(1/2) = -2That is, gradient of perpendicular = -2.This line goes through (0,3),(y - 3) = 2*(x - 0)y - 3 = 2xy = 2x + 3L1: y = 1/2*x - 3Gradient of the line = 1/2Negative reciprocal of gradient = -1/(1/2) = -2That is, gradient of perpendicular = -2.This line goes through (0,3),(y - 3) = 2*(x - 0)y - 3 = 2xy = 2x + 3
The gradient of a line is the same as the slope of a line. It will tell someone measuring the line how straight the line is.
The higher the gradient, the more steeper the line will be.
If you mean points of (1, 3) and (4, 3) then the gradient is 0 and it is a horizontal straight line parallel to the x axis on the Cartesian plane.
Yes, but the the line is vertical and the slope (or gradient) is undefined.
The gradient of a straight line cannot be defined- it's infinity.
A straight line with a gradient > 0 represents a constant rate of acceleration.