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What is the perpendicular bisector equation of the line joined by the points of h k and 3h -5k on the coordinated grid?
Wiki User
∙ 7y ago
8
Lorenza Kassulke
Wiki User
∙ 7y agoPoints: (h, k) and (3h, -5k)
Midpoint: (2h, -2k)
Slope: -3k/h
Perpendicular slope: h/3k
Perpendicular equation: y--2k = h/3k(x-2h) => 3ky--6K^2 = hx-2h^2
Perpendicular bisector equation: 3ky = hx -2h^2 -6K^2
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What is the perpendicular bisector equation of the line joined by the points of s 2s and 3s 8s?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular equation: y-5s = -1/3(x-2s) => 3y-15s = -x+2s => 3y = -x+17s Perpendicular bisector equation in its general form: x+3y-17s = 0
What is the perpendicular bisector equation of the line joined by the points of 7 3 and -6 1?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Equation: y-2 = -13/2(x-0.5) => 2y-4 = -13(x-0.5) => 2y = -13x+10.5 Perpendicular bisector equation in its general form: 13x+2y -10.5 = 0
What is the perpendicular bisector equation of the line joined by the points h k and 3h -5k showing work?
8
What is the perpendicular bisector equation of the line joined by the points of p q and 7p 3q on the Cartesian plane?
Points: (p, q) and (7p, 3q) Midpoint: (4p, 2q) Slope: q/3p Perpendicular slope: -3p/q Perpendicular bisector equation:- => y-2q = -3p/q(x-4p) => qy-2q^2 = -3p(x-4p) => qy-2q^2 = -3px+12p^2 => qy = -3px+12p^2+2q^2 In its general form: 3px+qy-12p^2-2q^2 = 0
What is the perpendicular bisector equation of the line joined by the points of p q and 7p 3q showing work and final answer in its general form?
The perpendicular bisector of a line has a gradient (m') which is the negative reciprocal of the gradient (m) of the line, and passes through the mid-point of the line.The equation of a line with gradient m through a point (x0, y0) has an equation of the form:y - y0 = m(x - x0)The gradient m of a line between two points (x0, y0) and (x1, y1) is given by:m = change_in_y / change_in_x = (y1 - y0) / (x1 - x0)Thus the line through (p, q) and (7p, 3q) has gradient:m = (3q - q) / (7p - p) = 2q / 6p = q/3pand the perpendicular bisector has gradient:m' = -1 / m = -1 / (q/3p) = -3p/qThe midpoint of the line through (p, q) and (3p, 3q) is:midpoint = ((p + 7p)/2), (q + 3q)/2) = (4p, 2q)Thus the perpendicular bisector of the line between (p, q) and (7p, 3q) has equation:y - 2q = -3p/q (x - 4p)→ qy - 2q² = -3px + 12p²→ qy + 3px = 12p² + 2q²Additional Information:-Final answer in its general form: 3px+qy-12p^2-2q^2 = 0
What is the perpendicular bisector equation of the line joined by the points -2 5 and -8 -3 on the Cartesian plane?
Points: (-2, 5) and (-8, -3) Midpoint: (-5, 1) Slope: 4/3 Perpendicular slope: -3/4 Perpendicular equation: y-1 = -3/4(x--5) => 4y = -3x-11 Perpendicular bisector equation in its general form: 3x+4y+11 = 0
What is the perpendicular bisector equation of the line joined by the points of -1 3 and -2 -5?
Points: (-1, 3) and (-2, -5) Midpoint: (-3/2, -1) Slope: 8 Perpendicular slope: -1/8 Perpendicular bisector equation: y--1 = -1/8(x--3/2) => y = -1/8x-19/16
What is the perpendicular bisector equation of the line segment joined by the points -2 4 and -4 8?
Points: (-2, 4) and (-4, 8) Midpoint: (-3, 6) Slope: -2 Perpendicular slope: 1/2 or 0.5 Perpendicular bisector equation: y-6 = 0.5(x--3) meaning y = 0.5x+7.5
What is the perpendicular bisector equation of the line joined by the points of s 2s and 3s 8s?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular equation: y-5s = -1/3(x-2s) => 3y-15s = -x+2s => 3y = -x+17s Perpendicular bisector equation in its general form: x+3y-17s = 0
What is the perpendicular bisector equation of the line joined by the points of 7 3 and -6 1?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Equation: y-2 = -13/2(x-0.5) => 2y-4 = -13(x-0.5) => 2y = -13x+10.5 Perpendicular bisector equation in its general form: 13x+2y -10.5 = 0
What is the perpendicular bisector equation of the line joined by the points h k and 3h -5k showing work?
8
What is the perpendicular bisector equation of the line joined by the points of p q and 7p 3q on the Cartesian plane?
Points: (p, q) and (7p, 3q) Midpoint: (4p, 2q) Slope: q/3p Perpendicular slope: -3p/q Perpendicular bisector equation:- => y-2q = -3p/q(x-4p) => qy-2q^2 = -3p(x-4p) => qy-2q^2 = -3px+12p^2 => qy = -3px+12p^2+2q^2 In its general form: 3px+qy-12p^2-2q^2 = 0
What is the perpendicular bisector equation of the line joined by the points of p q and 7p 3q showing work and final answer in its general form?
The perpendicular bisector of a line has a gradient (m') which is the negative reciprocal of the gradient (m) of the line, and passes through the mid-point of the line.The equation of a line with gradient m through a point (x0, y0) has an equation of the form:y - y0 = m(x - x0)The gradient m of a line between two points (x0, y0) and (x1, y1) is given by:m = change_in_y / change_in_x = (y1 - y0) / (x1 - x0)Thus the line through (p, q) and (7p, 3q) has gradient:m = (3q - q) / (7p - p) = 2q / 6p = q/3pand the perpendicular bisector has gradient:m' = -1 / m = -1 / (q/3p) = -3p/qThe midpoint of the line through (p, q) and (3p, 3q) is:midpoint = ((p + 7p)/2), (q + 3q)/2) = (4p, 2q)Thus the perpendicular bisector of the line between (p, q) and (7p, 3q) has equation:y - 2q = -3p/q (x - 4p)→ qy - 2q² = -3px + 12p²→ qy + 3px = 12p² + 2q²Additional Information:-Final answer in its general form: 3px+qy-12p^2-2q^2 = 0
What is the perpendicular bisector equation of a line joined by the points of s 2s and 3s 8s showing key stages of work?
It is found as follows:- Points: (s, 2s) and (3s, 8s) Slope: (2s-8s)/(s-3s) = -6s/-2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) Multiply all terms by 3: 3y-15s = -1(x-2s) => 3y = -x+17s In its general form: x+3y-17s = 0
What is the perpendicular bisector equation of the line joined by the points 11 13 and 17 19 showing work and answer?
The equation of a line through point (x0, y0) with gradient m is given by:y - y0 = m(x - x0)The gradient (m) of a line between two points (x0, y0) and (x1, y1) is given by:m = change_in_y/change_in_x = (y1 - y0)/(x1 - x0)→ The equation of the line between (11, 13) and (17, 19) is given by:y - 13 = (19-13)/(17-11) (x - 11)→ y - 13 = 6/6 (x - 11)→ y - 13 = x - 11→ y = x + 2and its gradient is m = 1.The gradient (m') of a line perpendicular to a line with gradient m is such that mm' = -1, ie m' = -1/m→ The gradient of the perpendicular line to the line between (11, 13) and (17, 19) has gradient m' = -1/1 = -1.The perpendicular bisector goes through the point midway between (11, 13) and (17, 19) which is given by the average of the x and y coordinates: ((11+17)/2, (13+19)/2) = (14, 16)Thus the perpendicular bisector of the line joining (11, 13) to (17,19) is given by:y - 16 = -1(x - 14)→ y - 16 = -x + 14→ y + x = 30Which in its general form is: x+y-30 = 0
Does an equilateral triangle have any perpendicular sides?
no a equilateral triangle does not have any perpendicular sides because the lines are joined together.
What word equation summarizes the dehydration of a carbohydrate?
monosaccharides become joined together
