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What is the perpendicular distance from the point 4 -2 to the line 2x -y -5 equals 0 showing key stages of work?
Wiki User
∙ 10y ago
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
Wiki User
∙ 10y ago
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What is the perpendicular distance from the point 7 5 to the staight line equation 3x plus 4y -16 equals 0 showing key stages of work?
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
What is the perpendicular distance from the coordinates of 7 and 5 to the straight line of 3x plus 4y -16 equals 0 showing key stages of work?
Points: (7, 5) Equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Equations intersect at: (4, 1) Length of perpendicular line: 5
What is the perpendicular distance from the point of 2 4 on the Cartesian plane to the straight line equation of y equals 2x plus 10 showing key stages of work?
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
What is the perpendicular distance from the point 2 and 4 to the straight line equation of y equals 2x plus 10 showing key stages of work?
1 Coordinates: (2, 4) 2 Equation: y = 2x+10 3 Perpendicular equation: y = -0.5+5 4 They intersect at: (-2, 6) 5 Distance is the square root of: (-2, -2)2+(6, -4) = 2*sq rt of 5 = 4.472 to 3 decimal places
A boat traveled 8 km east 5 km north then 2km west. How far is the boat from its starting point?
The total EW distance is 8 - 2 = 6 km.The total NS distance is 5 kmTherefore, total distance from the start is sqrt(62 + 52) = sqrt(36 + 25) = sqrt(61) =7.8 km (approx).The answer assumes that the boat is sufficiently far from the poles so that the various stages of its travel is along perpendicular lines.The total EW distance is 8 - 2 = 6 km.The total NS distance is 5 kmTherefore, total distance from the start is sqrt(62 + 52) = sqrt(36 + 25) = sqrt(61) =7.8 km (approx).The answer assumes that the boat is sufficiently far from the poles so that the various stages of its travel is along perpendicular lines.The total EW distance is 8 - 2 = 6 km.The total NS distance is 5 kmTherefore, total distance from the start is sqrt(62 + 52) = sqrt(36 + 25) = sqrt(61) =7.8 km (approx).The answer assumes that the boat is sufficiently far from the poles so that the various stages of its travel is along perpendicular lines.The total EW distance is 8 - 2 = 6 km.The total NS distance is 5 kmTherefore, total distance from the start is sqrt(62 + 52) = sqrt(36 + 25) = sqrt(61) =7.8 km (approx).The answer assumes that the boat is sufficiently far from the poles so that the various stages of its travel is along perpendicular lines.
What is the perpendicular bisector equation in its general form of the line whose coodinates are at s 2s and 3s 8s on the Cartesian grid showing key stages of work?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular bisector equation: y-5s = -1/3(x-2s) => 3y = -x+17s In its general form: x+3y-17s = 0
What is the length of the line and its perpendicular bisector equation that spans the points of 2 3 and 5 7 showing key stages of work?
Points: (2, 3) and (5, 7) Length of line: 5 Slope: 4/3 Perpendicular slope: -3/4 Midpoint: (3.5, 5) Bisector equation: 4y = -3x+30.5 or as 3x+4y-30.5 = 0
What are the solutions to the simultaneous equations of 6.6x plus 16.5y equals 52.8 and -16.5x -6.6y equals 6.6 showing key stages of work?
Eqn (A): => 2x + 5y = 16 Eqn (B): => 5x + 2y = -2 5*Eqn (A) - 2*Eqn (B): 21y = 84 => y = 4 Substituting for y in Eqn (a): x = -2
What is the perpedicular bisector equation of a line that cuts through the line of h plus k and 3h minus 5k showing key stages of work?
Points: (h, k) and (3h, -5k) Slope: -3k/h Perpendicular slope: h/3k Midpoint: (2h, -2k) Perpendicular equation: y--2k = h/3k(x-2h) Multiply all terms by 3k: 3ky--6k2 = h(x-2h) Equation in terms of 3ky = hx-2h2-6k2 Perpendicular bisector equation in its general form: hx-3ky-2h2-6k2 = 0
What in its general form is the perpendicular bisector equation passing through the line segment of h k and 3h -5k showing key stages of your work?
8
What is the perpendicular bisector equation of the line spanning the points of h k and 3h -5k on the Cartesian plane showing key stages of work?
8
What is a long distance biking event called?
It is called a tour. A tour have one stages or more
