Equation: y = 2x+10 and slope is 2
Perpendicular slope: -1/2
Perpendicular equation: 2y = -x+20
Both equations intersect at: (0, 10) from (4, 8)
Distance: square root of (0-4)^2 plus (10-8)^2 = 4.472 to three decimal places
If: 3y = 9x+18 then y = 3x+6 with a slope of 3 Perpendicular slope: -1/3 Perpendicular equation: y-29 = -1/3(x-19) => 3y = -x+106 Both equations intercept at: (8.8, 32.4) Perpendicular distance: square root of (8.8-19)^2+(32.4-29)^2 = 10.75 rounded
Points: (4, -6) and (2, -3) Slope: -3/2 Equation: 2y = -3x Perpendicular slope: 2/3 Perpendicular equation: 3y = 2x+3 Both equations meet at: (-6/13, 9/13) Distance from (6, 5) to (-6/13, 9/13) is 7.766 units rounded to 3 decimal places
Equation: 5x-2y = 3 Perpendicular equation: 2x+5y = -14 Both equations intersect at: (-13/29, -76/29) Perpendicular distance to 3 decimal places: 3.714
If you mean the perpendicular distance then it is worked out as follows:- Equation: y = 2x+10 Perpendicular slope: -1/2 Perpendicular equation: y-4 = -1/2(x-2) => 2y = -x+10 The two equations intersect at: (-2,6) Perpendicular distance is the square root of: (-2-2)2+(6-4)2 = 4.472 to 3 d.p.
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
If: 3y = 9x+18 then y = 3x+6 with a slope of 3 Perpendicular slope: -1/3 Perpendicular equation: y-29 = -1/3(x-19) => 3y = -x+106 Both equations intercept at: (8.8, 32.4) Perpendicular distance: square root of (8.8-19)^2+(32.4-29)^2 = 10.75 rounded
Points: (4, -6) and (2, -3) Slope: -3/2 Equation: 2y = -3x Perpendicular slope: 2/3 Perpendicular equation: 3y = 2x+3 Both equations meet at: (-6/13, 9/13) Distance from (6, 5) to (-6/13, 9/13) is 7.766 units rounded to 3 decimal places
The perpendicular distance from (2, 4) to the equation works out as the square root of 20 or 2 times the square root of 5
Equation: 5x-2y = 3 Perpendicular equation: 2x+5y = -14 Both equations intersect at: (-13/29, -76/29) Perpendicular distance to 3 decimal places: 3.714
Known equation: y = 2x+10 Perpendicular equation: 2y = -x+10 Both equations intersect at: (-2, 6) Distance from (2, 4) to (-2, 6) is sq rt of 20 using the distance formula
It works out as: 2 times the square root of 5
If you mean the perpendicular distance then it is worked out as follows:- Equation: y = 2x+10 Perpendicular slope: -1/2 Perpendicular equation: y-4 = -1/2(x-2) => 2y = -x+10 The two equations intersect at: (-2,6) Perpendicular distance is the square root of: (-2-2)2+(6-4)2 = 4.472 to 3 d.p.
the perpendicular distance from the base of a quadrilateral to the opposite side?
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: y = 2x+10 Point: (2, 4) Perpendicular slope: -1/2 Perpendicular equation: y-4 = -1/2(x-2) => 2y = -x+10 Both equations intersect at: (-2, 6) Using distance formula: (2, 4) to (-2, 6) = 2 times square root of 5
Equation: 3x+4y = 0 => y = -3/4x Perpendicular slope: 4/3 Perpendicular equation: 4x-3y-13 = 0 Equations intersect at: (2.08, -1.56) Distance from (7, 5) to (2.08, -1.56) = 8.2 units using the distance formula
Perpendicular distance is used while calculating movement in order to verify the calculation. The perpendicular calculation must match the original calculation.