Chord equation: y = x+5
Circle equation: x^2 +4x +y^2 -18y +59 = 0
Chord end points: (-1, 4) and (3, 8)
Chord midpoint: (1, 6)
Perpendicular slope: -1
Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
If: y = 5x +10 and y = x^2 +4 Then: x^2 +4 = 5x +10 Transposing terms: x^2 -5x -6 = 0 Factorizing the above: (x-6)(X+1) = 0 meaning x = 6 or x = -1 Therefore by substitution endpoints of the line are at: (6, 40) and (-1, 5) Midpoint of line: (2.5, 22.5) Slope of line: 5 Perpendicular slope: -1/5 Perpendicular bisector equation: y-22.5 = -1/5(x-2.25) => 5y = -x+114.75 Perpendicular bisector equation in its general form: x+5y-114.75 = 0
Points: (4, 1) and (0, 4) Slope: -3/4 Equation: 4y = -3x+16 Perpendicular slope: 4/3 Perpendicular equation: 3y = 4x-13 Both equations meet at: (4, 1) from (7, 5) at right angles Perpendicular distance: square root of [(4-7)squared+(1-5)squared)] = 5 units
Equation: 7x+13-y = 0 => y = 7x+13Slope: 7Perpendicular slope: -1/7Perpendicular equation: 7y = -x+37Both equations intersect at: (-1.08, 5.44)Distance: square root of [(-1.08-2)squared+(5.44-5)squared)] = 3.111 to 3 d.p.
no
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
If: y = 5x +10 and y = x^2 +4 Then: x^2 +4 = 5x +10 Transposing terms: x^2 -5x -6 = 0 Factorizing the above: (x-6)(X+1) = 0 meaning x = 6 or x = -1 Therefore by substitution endpoints of the line are at: (6, 40) and (-1, 5) Midpoint of line: (2.5, 22.5) Slope of line: 5 Perpendicular slope: -1/5 Perpendicular bisector equation: y-22.5 = -1/5(x-2.25) => 5y = -x+114.75 Perpendicular bisector equation in its general form: x+5y-114.75 = 0
Points: (4, 1) and (0, 4) Slope: -3/4 Equation: 4y = -3x+16 Perpendicular slope: 4/3 Perpendicular equation: 3y = 4x-13 Both equations meet at: (4, 1) from (7, 5) at right angles Perpendicular distance: square root of [(4-7)squared+(1-5)squared)] = 5 units
Equation: 7x+13-y = 0 => y = 7x+13Slope: 7Perpendicular slope: -1/7Perpendicular equation: 7y = -x+37Both equations intersect at: (-1.08, 5.44)Distance: square root of [(-1.08-2)squared+(5.44-5)squared)] = 3.111 to 3 d.p.
no
A quadratic equation.
the name is squared equation
Sin squared, cos squared...you removed the x in the equation.
No, It's a a quadratic equation because you have X squared.
11
A squared plus b squared equils c squared
x2 = 657.4096