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What is the equation of a circle whose diameter endpoints are at 10 -4 and 2 2 on the Cartesian plane and what are the perpendicular equations to its endpoints showing work?
Endpoints of diameter: (10, -4) and (2, 2)Midpoint which is the center of the circle: (6, -1)Distance from (10, -4) or (2, 2) to (6, -1) = 5 which is the radius of the circleEquation of the circle: (x-6)^2 +(y+1)^2 = 25Slope of radius: -3/4Slope of perpendicular equations which will be parallel: 4/31st perpendicular equation: y--4 = 4/3(x-10) => 3y = 4x-522nd perpendicular equation: y-2 = 4/3(x-2) => 3y = 4x-2
An equation showing that two ratios are equal?
A proportion.
What is the equation of a circle whose endpoints of its diameter are at 2 2 and 10 -4 on the Cartesian plane showing work?
Endpoints: (2, 2) and (10, -4) Midpoint: (6, -1) which is the centre of the circle Distance from (6, -1) to (2, 2) or (10, -4) = 5 which is the radius of the circle Therefore equation of the circle: (x-6)^2 + (y+1)^2 = 25
What is the perpendicular distance from the coordinate of 2 5 that meets the staight line of y equals 7x plus 13 at right angles on the Cartesian plane showing all work and answer to 3 decimal places?
Coodinate: (2, 5) Equation: y = 7x+13 Slope: 7 Perpendicular slope: -1/7 Perpendicular equation: 7y = -x+37 Both equations intersect at: (-1.08, 5.44) Perpendicular distance: square root of [(-1.08-2)^2+(5.44-5)^2] = 3.111 to 3 d.p.
What is the perpendicular bisector equation of the line joining the points 7 3 and -6 1 on the Cartesian plane showing work?
Let P = (7, 3) and Q = (-6, 1) Then mid point of PQ = ((7-6)/2, (3+1)/2) = (1/2, 2) Also, gradient of PQ = (3 - 1)/(7 + 6) = 2/13 So the gradient of the perpendicular = -13/2 Therefore the required line passes through the point (1/2, 2) and has gradient -13/2 and so its equation is (y - 2) = -13/2*(x - 1/2) or 4y - 8 = -26x + 13 that is 26x + 4y - 21 = 0
What is the perpendicular bisector equation that meets the line 13 19 and 23 17 at midpoint on the Cartesian plane showing all aspects of work with answer?
Points: (13, 19) and (23, 17) Midpoint: (18, 18) Slope: -1/5 Perpendicular slope: 5 Perpendicular equation: y-18 = 5(x-18) => y = 5x-72
What is the perpendicular bisector equation that meets the line segment of -2 2 and 6 4 at its midpoint showing work?
Points: (-2, 2) and (6, 4) Midpoint: (2, 3) Slope: 1/4 Perpendicular slope: -4 Perpendicular bisector equation: y-3 = -4(x-2) => y = -4x+11
What is the perpendicular equation that meets the line of 7 3 and -6 1 at its midpoint showing work?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Perpendicular equation: y-2 = -13/2(x-0.5) => 2y-4 = -13x+6.5 => 2y = -13x+10.5
What is the perpendicular equation in its general form that meets the line whose coordinates are 2 5 and 11 17 at its midpoint on the Cartesian plane showing all work?
Points: (2, 5) and (11, 17) Midpoint: (6.5, 11) Slope: 4/3 Perpendicular slope: -3/4 Perpendicular equation: y-11 = -3/4(x-6.5) => 4y = -3x+63.5 In its general form: 3x+4y-63.5 = 0
What is the perpendicular bisector equation that meets the line segment of 7 3 and -6 1 showing work in addition to the answer?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Perpendicular bisector equation: y-2 = -13/2(x-0.5) => 2y = -13x+10.5
What is the perpendicular bisector equation in its general form that meets the line containing the points 7 3 and -6 1 showing work?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Perpendicular equation: y-2 = -13/2(x-0.5) => 2y = -13x+10.5 Perpendicular bisector equation in its general form: 26x+4y-21 = 0
What is the equation of a straight line that cuts through the middle of the points of -1 3 and -2 -5 at right angles on the Cartesian plane showing work?
The equation will be a perpendicular bisector equation of the given points:- Points: (-1, 3) and (-2, -5) Midpoint: (-3/2, -1) Slope: 8 Perpendicular slope: -1/8 Perpendicular equation: y--1 = -1/8(x--3/2) => y = -1/8x-3/16-1 Therefore the perpendicular bisector equation is: y = -1/8x -19/16
What is the equation and its perpendicular distance from the point -3 -5 whose line meets the line of 0 5 to 3 0 at its midpoint on the Cartesian plane showing work?
Points: (0, 5) and (3, 0) Midpoint: (1.5, 2.5) Slope: -5/3 Perpendicular slope: 3/5 Perpendicular equation: y--5 = 3/5(x--3) => 5y = 3x-16 Distance is the square root of (1.5--3)^2+(2.5--5)^2 = 8.746 to three decimal places
What is the perpendicular bisector equation of the line joined by the points h k and 3h -5k showing work?
8
What is the length of the line and its perpendicular bisector equation that spans the points of 2 3 and 5 7 showing key stages of work?
Points: (2, 3) and (5, 7) Length of line: 5 Slope: 4/3 Perpendicular slope: -3/4 Midpoint: (3.5, 5) Bisector equation: 4y = -3x+30.5 or as 3x+4y-30.5 = 0
What is the perpendicular bisector equation in its general form of the line whose coodinates are at s 2s and 3s 8s on the Cartesian grid showing key stages of work?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular bisector equation: y-5s = -1/3(x-2s) => 3y = -x+17s In its general form: x+3y-17s = 0
What is the perpedicular bisector equation of a line that cuts through the line of h plus k and 3h minus 5k showing key stages of work?
Points: (h, k) and (3h, -5k) Slope: -3k/h Perpendicular slope: h/3k Midpoint: (2h, -2k) Perpendicular equation: y--2k = h/3k(x-2h) Multiply all terms by 3k: 3ky--6k2 = h(x-2h) Equation in terms of 3ky = hx-2h2-6k2 Perpendicular bisector equation in its general form: hx-3ky-2h2-6k2 = 0
