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 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
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
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
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
Points: (0, 5) and (3, 0) Midpoint: (1.5, 2.5) Slope: -5/3 Perpendicular slope: 3/5 Perpendicular equation: y--5 = 3/5(x--3) => 5y = 3x-16 Distance is the square root of (1.5--3)^2+(2.5--5)^2 = 8.746 to three decimal places
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
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
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
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
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
Points: (0, 5) and (3, 0) Midpoint: (1.5, 2.5) Slope: -5/3 Perpendicular slope: 3/5 Perpendicular equation: y--5 = 3/5(x--3) => 5y = 3x-16 Distance is the square root of (1.5--3)^2+(2.5--5)^2 = 8.746 to three decimal places
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
Coordinate: (11, 17)If: 3x+4y-63.5 = 0Then: 4y = -3x+63.5 => y = -3/4x+15.875Slope: -3/4Perpendicular slope: 4/3Perpendicular equation: y-17 = 4/3(x-11) => 3y = 4x+7Both equations intersect at: (6.5, 11)Perpendicular distance: square root of [(6.5-11)^2+(11-17)^2] = 7.5
Points: (4, -2) Equation: 2x-y-5 = 0 Perpendicular equation: x+2y = 0 Equations intersect at: (2, -1) Perpendicular distance is the square root of: (2-4)2+(-1--2)2 = 5 Distance = square root of 5
Coodinate: (2, 5) Equation: y = 7x+13 Slope: 7 Perpendicular slope: -1/7 Perpendicular equation: 7y = -x+37 Both equations intersect at: (-1.08, 5.44) Perpendicular distance: square root of [(-1.08-2)^2+(5.44-5)^2] = 3.111 to 3 d.p.
Equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Both equations intersect at: (4, 1) Perpendicular distance: square root of (7-4)2+(5-1)2 = 5