3x-4y-6 = 0
y=-x
Y=4x+3
A binormal plane is the straight line passing through a point M0 of a curve L perpendicular to the oscillating plane to L at M0. If r=r(t) is a parameterization of L, then the vector equation of the binormal at M0 corresponding to the value of t0 to the parameter of t has the form.
2x-y -5 = 0 => y = 2x -5 The perpendicular slope is the negative reciprocal of 2 which is -1/2 So using the formula of y -y1 = m(x -x1) gives the straight line equation:- y - -2 = -1/2(x -4) y +2 = -1/2x +2 y = -1/2x +2 -2 y = -1/2x which can be expressed in the form of: x +2y = 0 So the straight line equation is: x +2y = 0
Given a straight line joining the points A and B, the perpendicular bisector is a straight line that passes through the mid-point of AB and is perpendicular to AB.
7x-y-28 = 0
y=-x
It is the equation of a straight line in the form of: y = 2x+4
5x - 10 = -20This equation can be restated as 5x = -10 : x = -2This is the equation of a straight line perpendicular to the x axis and passing through the point x = -2. There is no y intercept and the slope is indeterminate.
There is no name for it except "A line perpendicular to a line segment and passing through its midpoint".
Known equation: 5x -2y = 3 or y = 5/2x -3/2 Slope of equation: 5/2 Slope of perpendicular equation: -2/5 Perpendicular equation: y --4 = -2/5(x -3) => 5y = -2x -14 Perpendicular equation in its general form: 2x+5y+14 = 0
Without an equality sign and not knowing the plus or minus values of y and 7 it can't be considered to be a straight line equation therefore finding its perpendicular equation is impossible.
The equation will be perpendicular to the given equation and have a slope of 3/4:- Perpendicular equation: y--3 = 3/4(x--2) => 4y--12 = 3x--6 => 4y = 3x-6 Perpendicular equation in its general form: 3x-4y-6 = 0
y = 2.25x -6
Known equation: 5x-2y = 3 or y = 5/2x -3/2 Slope of known equation: 5/2 Slope of perpendicular equation: -2/5 Perpendicular equation: y- -4 = -2/5(x-3) => 5y =-2x-14 Perpendicular equation in its general form: 2x+5y+14 = 0
2y+x+5 = 0
What is the equation of the vertical line passing through (-5,-2)