Points of line: (13, 17) and (19, 23)
Its slope: 1
Its equation: y = x+4 => y-x = 4
Multiply all terms by 4: 4y-4x = 16
Equation of: 4y = 5x => 4y-5x = 0
Subtacting equations: x = 16
By substitution point of intersection is at: (16, 20)
Points: (13, 17) and (19, 23) Midpoint: (16, 20) Slope of required equation: 5/4 Its equation: 4y = 5x or as y = 1.25x Its distance from (0, 0) to (16, 20) = 4 times sq rt 41
1 Equation: y = 2x+10 2 Perpendicular equation works out as: 2y = -x+10 3 Point of intersection: (-2, 6) 4 Distance is the square root of: (-2-2)2+(6-4)2 = 2 times sq rt of 5
Points: (2, 3) and (11, 13) Slope: 10/9 Equation: 9y = 10x+7
Improved Answer:-If: 2x+y = 5 and x^2 -y^2 = 3Then by rearranging: y = 5 -2x and -3x^2 -28+20x = 0Solving the above quadratic equation: x = 2 and x = 14/3By substitution points of intersection are: (2, 1) and (14/3, -13/3)
Points: (2, 3) and (5, 7)Length: 5 unitsSlope: 4/3Perpendicular slope: -3/4Midpoint: (3.5, 5)Equation: 3y = 4x+1Bisector equation: 4y = -3x+30.5
Points: (s, 2s) and (3s, 8s) Slope: (8s-2s)/(3s-s) = 6s/2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) => 3y-15s = -1(x-2s) => 3y = -x+17x Perpendicular bisector equation in its general form: x+3y-17s = 0
They must be equidistant from the point of bisection which is their midpoint and works out that a = -2 and b = 4 Sketching the equations on the Cartesian plane will also help you in determining their values
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
A proportion.
PROPORTION
A balanced equation is representative for the amounts and nature for reactants and products involved.
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