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What is a line perpendicular to a line segmentt and passing through its midpoint?
There is no name for it except "A line perpendicular to a line segment and passing through its midpoint".
What is the shortest path between a line and a point not on that line?
The shortest path is a line perpendicular to the given line that passes through the given point.
What is the general form of a straight line equation perpendicular to 4x plus 3y minus 5 equals 0 passing through a point minus 2 and minus 3 on the line?
Equation of original line is 4x + 3y - 5 = 0 that is, 3y = -4x + 5 or y = -(4/3)x + 5 Slope of original line = -4/3 Slope of line perpendicular to it = 3/4 General equation of perpendicular line: y = (3/4)x + c for some constant c or 4y = 3x + c' The point (-2,-3) is on this line so 4*(-3) = 3*(-2) + c' -12 = - 6 + c' so that c' = -6 The equation of the perpendicular line is 4y = 3x - 6
Is the line through points p0 9 and q2 -8 perpendicular to the line through points?
No, because the second line is not defined.
Which line is perpendicular to 2x-8y plus 23 equals 0?
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
