Q: What is rhe perpendicular distance from the point 4 8 to the line of y equals 2x plus 10 on the Cartesian plane showing key aspects of work with an answer to an appropriate degree of accuracy?

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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

Points: (4, -6) and (2, -3) Slope: -3/2 Equation: 2y = -3x Perpendicular slope: 2/3 Perpendicular equation: 3y = 2x+3 Both equations meet at: (-6/13, 9/13) Distance from (6, 5) to (-6/13, 9/13) is 7.766 units rounded to 3 decimal places

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

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.

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

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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

Points: (4, -6) and (2, -3) Slope: -3/2 Equation: 2y = -3x Perpendicular slope: 2/3 Perpendicular equation: 3y = 2x+3 Both equations meet at: (-6/13, 9/13) Distance from (6, 5) to (-6/13, 9/13) is 7.766 units rounded to 3 decimal places

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

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

It works out as: 2 times the square root of 5

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.

the perpendicular distance from the base of a quadrilateral to the opposite side?

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 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

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

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