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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

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Q: What is the distance from the point 5 7 that is perpendicular to to the straight line equation of 3x-y plus 2 equals 0 on the Cartesian plane showing key aspects of work?
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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 of 7 and 5 that meets the straight line equation of 3x plus 4y equals 0 on the Cartesian plane showing work?

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


What is the perpendicular distance from the point of 7 5 to the straight line whose equation is 3x plus 4y minus 16 equals 0?

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


What is the perpendicular distance from the point 4 -2 to the straight line of 2x -y -5 equals 0?

Perpendicular equation: x+2y = 0 Point of intersection: (2, -1) Perpendicular distance: square root of 5


What is the perpendicular distance from the point 7 and 5 to the straight line equation of 3x plus 4y minus 16 equals 0?

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

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What is the perpendicular distance from the point of 2 4 to the straight line of y equals 2x plus 10 on the Cartesian plane?

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


What is the perpendicular distance from the point 2 4 to the straight line equation of y equals 2x plus 10 on the Cartesian plane?

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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 of 7 and 5 that meets the straight line equation of 3x plus 4y equals 0 on the Cartesian plane showing work?

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


What is the perpendicular distance from the point of 7 5 to the straight line whose equation is 3x plus 4y minus 16 equals 0?

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


What is the perpendicular distance from the point 4 -2 to the straight line of 2x -y -5 equals 0?

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What is the perpendicular distance from the point 7 and 5 to the straight line equation of 3x plus 4y minus 16 equals 0?

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


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