Write the equation of the given line in standard form: y = mx + c
so: 3y = 4x - 7 or y = 4/3x - 7/3
The slope of the given line is 4/3.
The slope of the line perpendicular to it is the negative reciprocal (change the sign and flip the fraction over).
ie the perpendicular has slope -3/4
So the equation of the perpendicular is of the form y = -3/4x + d
or 4y = - 3x + 4d
or 3x + 4y = 4d.
To evaluate the value of the constant d (or 4d) you will need one point on the perpendicular line. Substitute the coordinates of that point for x and y in the above equation and you've got 4d.
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
It is -1/2
7x + 10y = 4.5 : 10y = -7x + 4.5 : y = -x.7/10 + 0.45, the gradient of this line is -7/10 Two straight lines are perpendicular if the product of their gradients is -1. Let the equation for the perpendicular line be y = mx + c Then m x -7/10 = -1 : m = 10/7 The equation for the perpendicular line is y = x.10/7 + c If the values of x and y for the point of intersection are provided then these can be substituted in the perpendicular line equation and the value of c obtained. If appropriate, the equation can then be restructured to a format similar to the original equation.
Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point: (7, 5) Equations intersect: (4, 1) Perpendicular distance: square root of [(7-4)2+(5-1)2] = 5
Slope for the equation y equals 7 is zero.
12
-5
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
It is -1/2
It is: y = -1/5x+7 that is perpendicular to y = 5x+7
7x + 10y = 4.5 : 10y = -7x + 4.5 : y = -x.7/10 + 0.45, the gradient of this line is -7/10 Two straight lines are perpendicular if the product of their gradients is -1. Let the equation for the perpendicular line be y = mx + c Then m x -7/10 = -1 : m = 10/7 The equation for the perpendicular line is y = x.10/7 + c If the values of x and y for the point of intersection are provided then these can be substituted in the perpendicular line equation and the value of c obtained. If appropriate, the equation can then be restructured to a format similar to the original equation.
Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point: (7, 5) Equations intersect: (4, 1) Perpendicular distance: square root of [(7-4)2+(5-1)2] = 5
Equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Both equations intersect at: (4, 1) Perpendicular distance: square root of (7-4)2+(5-1)2 = 5
Points: (7, 5) Equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Equations intersect at: (4, 1) Length of perpendicular line: 5
4x+5y = 7 5y = -4x+7 y = -4/5x+7/5 in slope intercept form So the slope of the perpendicular line is plus 5/4
If the original line has the equation 2y = 7x then this can be written as y = 3.5x and the slope of this line is thus 3.5. Two straight lines are perpendicular if the product of their gradients is -1. Let m be the slope of the perpendicular line then, 3.5m = -1 : m = -1/3.5 = -2/7 The equation of the perpendicular line is thus, y = -2/7 * x + b where b is the intercept. As b = 5 then the equation becomes y = -2/7*x + 5 We can multiply by 7 to eliminate the fraction, 7y = -2x + 35 and re-order the terms to become 7y = 35 - 2x
Slope for the equation y equals 7 is zero.