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What in its general form is the perpendicular bisector equation passing through the line segment of h k and 3h -5k showing key stages of your work?
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What is the perpendicular bisector equation passing between the points 3 -4 and -1 -2?
Points: (3,-4) and (-1, -2) Midpoint: (1,-3) Slope: -1/2 Perpendicular slope: 2 Perpendicular bisector equation in slope intercept form: y = 2x-5
What is the perpendicular bisector equation passing through the line segment of 7 7 and 3 5 giving brief details?
Points: (7, 7) and (3, 5) Midpoint: (5, 6) Slope: 1/2 Perpendicular slope: -2 Use: y-6 = -2(x-5) Perpendicular bisector equation: y = -2x+16 or as 2x+y-16 = 0
What is the equation of the line passing through 3 -4 and it is perpendicular to the line 5x -2y equals 3?
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
What is the equation of the line passing through -2 -3 that is at right angles to the equation 4x plus 3y -5 equals 0 on the Cartesian plane?
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
What is the equation passing through the point 3 -4 which is perpendicular to the line 5x -2y equals 3?
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
How do you form an equation for the perpendicular bisector of the line segment joining -2 5 to -8 -3?
First find the midpoint of (-2, 5) and (-8, -3) which is (-5, 1) Then find the slope of (-2, 5) and (-8, -3) which is 4/3 Slope of the perpendicular bisector is the negative reciprocal of 4/3 which is -3/4 Now form an equation of the straight line with a slope of -3/4 passing through the point (-5, 1) using the formula y-y1 = m(x-x1) The equation works out as: 3x+4y+11 = 0
What is the perpendicular bisector equation of the line segment whose end points are at 7 3 and -6 1?
To solve this, four steps are needed:Find the midpoint of the line segment (X, Y) which is a point on the perpendicular bisectorFind the slope m for the line segment: m = change_in_y/change_in_xFind the slope m' of the perpendicular line; the slopes of the lines are related by mm' = -1 → m' = -1/mFind the equation of the perpendicular bisector using the slope-point equation for a line passing through point (X, Y) with slope m': y - Y = m'(x - X)Have a go before reading the solution below.--------------------------------------------------------------------The midpoint of (7, 3) and (-6, 1) is at ((7 + -6)/2, (3 + 1)/2) = (1/2, 2)The slope of the line segment is: m = change_in_y/change_in_x = (1 - 3)/(-6 - 7) = -2/-13 = 2/13The slope of the perpendicular bisector is m' = -1/m = -1/(2/13) = -13/2The equation of the perpendicular bisector passing through point (X, Y) = (1/2, 2) with slope m' = -13/2 is given by:y - Y = m'(x - Y)→ y - 2 = -13/2(x - 1/2)→ 4y - 8 = -26x + 13→ 4y + 26x = 21
What equation represents a line perpendicular to y equals x and passing through the points 0 0?
y=-x
What is the proof that an angle bisector actually bisects an angle?
In general 'to bisect' something means to cut it into two equal parts. The 'bisector' is the thing doing the cutting.In an angle bisector, it is a line passing through the vertex of the angle that cuts it into two equal smaller angles.Therefore it's in the definition.
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
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 equation that represents the line perpendicular to y equals -3x plus 4 and passing through the point -1 1?
y = 1/3x+4/3
