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
If you mean the perpendicular distance from the coordinate of (7, 5) to the straight line 3x+4y-16 = 0 then it works out as 5 units.
Equation of line: 3x+4y-16 = 0 or y = -3/4x+4 Slope of line: -3/4 Slope of perpendicular line: 4/3 Perpendicular equation: y-5 = 4/3(x-7) or 3y = 4x-13 Both equations intersect at: (4, 1) Perpendicular distance from (7, 5) to (4, 1) = 5 units using distance formula
The slope of the perpendicular is -(1/2) .
+ 1/8
"From the geometric point of view, the given point is the focus of the parabola and the given line is its directrix. It can be shown that the line of symmetry of the parabola is the line perpendicular to the directrix through the focus. The vertex of the parabola is the point of the parabola that is closest to both the focus and directrix."-http://www.personal.kent.edu/~rmuhamma/Algorithms/MyAlgorithms/parabola.htm"A line perpendicular to the axis of symmetry used in the definition of a parabola. A parabola is defined as follows: For a given point, called the focus, and a given line not through the focus, called the directrix, a parabola is the locus, or set of points, such that the distance to the focus equals the distance to the directrix."-http://www.mathwords.com/d/directrix_parabola.htm
It works out as: 2 times the square root of 5
Perpendicular equation: x+2y = 0 Point of intersection: (2, -1) Perpendicular distance: square root of 5
The x and y axes on the Cartesian plane are perpendicular to each other at the point of origin
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
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
If you mean the perpendicular distance from the coordinate of (7, 5) to the straight line 3x+4y-16 = 0 then it works out as 5 units.
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
It cannot.
Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point of intersection: (4, 1) Distance: (7-4)2+(5-1)2 = 25 and the square root of this is the perpendicular distance which is 5 units of measurement
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.
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