To find the perpendicular distance from a given point to a given line, find the equation of the line perpendicular to the given line which passes through the given point. Then the distance can be calculated as the distance from the given point to the point of intersection of the two lines, which can be calculated by using Pythagoras on the Cartesian coordinates of the two points.
A line in the form y = mx + c has gradient m.
If a line has gradient m, the line perpendicular to it has gradient m' such that mm' = -1, ie m' = -1/m (the negative reciprocal of the gradient).
A line through a point (x0, y0) with gradient m has equation:
y - yo = m(x - x0)
Thus the equation of the line through (5, 7) that is perpendicular to 3x - y + 2 = 0 can be found.
The intercept of this line with 3x - y + 2 = 0 can be calculated as there are now two simultaneous equations.
→ The perpendicular distance from (5, 7) to the line 3x - y + 2 = 0 is the distance form (5, 7) to this point of interception, calculated via Pythagoras:
distance = √((change_in_x)^2 + (change_in_y)^2)
This works out to be √10 ≈ 3.162
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
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
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
Perpendicular equation: x+2y = 0 Point of intersection: (2, -1) Perpendicular distance: square root of 5
Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point: (7, 5) Equations intersect: (4, 1) Perpendicular distance: square root of [(7-4)2+(5-1)2] = 5
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
It works out as: 2 times the square root of 5
A straight line on the Cartesian plane is the graph of a linear equation.
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
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
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
Perpendicular equation: x+2y = 0 Point of intersection: (2, -1) Perpendicular distance: square root of 5
Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point: (7, 5) Equations intersect: (4, 1) Perpendicular distance: square root of [(7-4)2+(5-1)2] = 5
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
The graph, in the Cartesian plane, of a linear equation is a straight line. Conversely, a straight line in a Cartesian plane can be represented algebraically as a linear equation. They are the algebraic or geometric equivalents of the same thing.
Plot its straight line equation on the Cartesian plane
It is the equation of a straight line plotted on the Cartesian plane.